Revision notes, practice numericals, a question bank, assertion and reason questions and answers to the NCERT questions. Tap a group to open it; opening one closes the others.
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A car travels 60 km east in 1 h and then 30 km west in the next 0.5 h. Find the average velocity and the average speed for the whole journey.
Given: East is positive: Δx₁ = +60 km in 1 h; Δx₂ = −30 km in 0.5 h
Use: Average velocity = displacement / time; average speed = path length / time
Displacement = 60 − 30 = +30 km; time = 1.5 h; average velocity = 30/1.5 = 20 km h⁻¹ (east).
Path length = 60 + 30 = 90 km; average speed = 90/1.5 = 60 km h⁻¹.
The average speed is larger than the magnitude of the average velocity.
Answer: 20 km h⁻¹ east; 60 km h⁻¹
The position of a particle is x = 4 + 6t² − t³ (x in metres, t in seconds). Find (a) the velocity at t = 2 s, (b) the time at which it is momentarily at rest, (c) the average velocity from t = 0 to t = 4 s.
Given: x = 4 + 6t² − t³
Use: v = dx/dt
v = dx/dt = 12t − 3t² (m s⁻¹).
(a) At t = 2 s: v = 24 − 12 = 12 m s⁻¹.
(b) v = 0: 3t(4 − t) = 0, so t = 0 or t = 4 s. The particle is momentarily at rest at t = 4 s (and at t = 0).
(c) x(0) = 4 m, x(4) = 4 + 96 − 64 = 36 m. Average velocity = (36 − 4)/4 = 8 m s⁻¹.
Answer: (a) 12 m s⁻¹; (b) 4 s; (c) 8 m s⁻¹
For x = 0.08 t³ (metres, seconds) find the average velocity between t = 3.5 s and t = 4.5 s and compare it with the instantaneous velocity at t = 4 s.
Given: x = 0.08 t³
Use: Average velocity = Δx/Δt; instantaneous velocity = dx/dt
x(4.5) = 0.08 × 91.125 = 7.29 m; x(3.5) = 0.08 × 42.875 = 3.43 m.
Average velocity = (7.29 − 3.43)/1.0 = 3.86 m s⁻¹.
v = dx/dt = 0.24 t² = 0.24 × 16 = 3.84 m s⁻¹ at t = 4 s.
The chord over a shorter interval would give a value even closer to 3.84 m s⁻¹.
Answer: 3.86 m s⁻¹ (average) and 3.84 m s⁻¹ (instantaneous)
A student walks 1.2 km to school at 4 km h⁻¹ and returns by the same road at 6 km h⁻¹. Find the average speed and the average velocity for the round trip.
Given: d = 1.2 km each way; v₁ = 4 km h⁻¹; v₂ = 6 km h⁻¹
Use: Average speed = total path / total time
Time going = 1.2/4 = 0.3 h; time returning = 1.2/6 = 0.2 h; total time = 0.5 h.
Total path = 2.4 km; average speed = 2.4/0.5 = 4.8 km h⁻¹ (not (4 + 6)/2 = 5).
The displacement is zero, so the average velocity is 0.
Answer: 4.8 km h⁻¹; 0
A bus covers the first half of a route at 40 km h⁻¹ and the second half at 60 km h⁻¹. Find the average speed for the whole route.
Given: Half-distances equal, d each; v₁ = 40, v₂ = 60 km h⁻¹
Use: Total path 2d divided by total time d/v₁ + d/v₂
Average speed = 2d / (d/40 + d/60) = 2/(1/40 + 1/60) = 2/(5/120) = 48 km h⁻¹.
This is the harmonic mean 2v₁v₂/(v₁ + v₂), not the arithmetic mean 50 km h⁻¹.
Answer: 48 km h⁻¹
The position of a particle is x = 5t² (m). Find its velocity at t = 3 s, and its average velocity between t = 2 s and t = 4 s. Comment on the result.
Given: x = 5t²
Use: v = dx/dt
v = dx/dt = 10t, so v(3) = 30 m s⁻¹.
x(2) = 20 m, x(4) = 80 m; average velocity = (80 − 20)/2 = 30 m s⁻¹.
They are equal because the acceleration is constant: the average velocity over an interval equals the instantaneous velocity at the middle of the interval.
Answer: 30 m s⁻¹ in both cases
A car's speed increases from 18 km h⁻¹ to 72 km h⁻¹ in 5 s along a straight road. Find the average acceleration.
Given: v₁ = 18 km h⁻¹ = 5 m s⁻¹; v₂ = 72 km h⁻¹ = 20 m s⁻¹; Δt = 5 s
Use: ā = Δv/Δt
ā = (20 − 5)/5 = 3 m s⁻².
Answer: 3 m s⁻²
The velocity of a particle is v = 3t² − 12 (m s⁻¹). Find the acceleration at t = 2 s and the instant when the particle is momentarily at rest.
Given: v = 3t² − 12
Use: a = dv/dt
a = dv/dt = 6t, so a(2) = 12 m s⁻².
v = 0: 3t² = 12, t = 2 s (taking t > 0).
At the instant the particle is at rest, its acceleration is not zero: 12 m s⁻².
Answer: 12 m s⁻²; at t = 2 s
A train starts from rest and its speed rises uniformly to 10 m s⁻¹ in 5 s, stays constant for 10 s and then falls uniformly to zero in 5 s. Find the total distance and the average speed.
Given: Three stages: 0–5 s, 5–15 s, 15–20 s
Use: Distance = area under the v–t graph
Stage 1 (triangle): ½ × 5 × 10 = 25 m.
Stage 2 (rectangle): 10 × 10 = 100 m.
Stage 3 (triangle): ½ × 5 × 10 = 25 m.
Total distance = 150 m in 20 s; average speed = 150/20 = 7.5 m s⁻¹.
Answer: 150 m; 7.5 m s⁻¹
A particle moves with velocity +4 m s⁻¹ for 3 s and then with velocity −2 m s⁻¹ for 5 s. Find its displacement and the distance covered.
Given: v = +4 m s⁻¹ (3 s); v = −2 m s⁻¹ (5 s)
Use: Signed area gives displacement; total area gives distance
Displacement = 4 × 3 + (−2) × 5 = 12 − 10 = +2 m.
Distance = 12 + 10 = 22 m.
The final position is 2 m from the start, in the positive direction.
Answer: Displacement +2 m; distance 22 m
The velocity of a body changes uniformly from +8 m s⁻¹ to −4 m s⁻¹ in 3 s. Find the acceleration and the instant at which the body is momentarily at rest.
Given: v₀ = +8 m s⁻¹; v = −4 m s⁻¹ after 3 s
Use: a = (v − v₀)/t and v = v₀ + at
a = (−4 − 8)/3 = −4 m s⁻².
v = 0: 8 − 4t = 0, t = 2 s.
The speed first decreases to zero (0–2 s) and then increases in the opposite direction (2–3 s).
Answer: −4 m s⁻²; at t = 2 s
A car starts from rest with a constant acceleration of 2 m s⁻². Find its speed and distance covered after 10 s.
Given: v₀ = 0; a = 2 m s⁻²; t = 10 s
Use: v = v₀ + at, x = v₀t + ½at²
v = 0 + 2 × 10 = 20 m s⁻¹.
x = ½ × 2 × 10² = 100 m.
Answer: 20 m s⁻¹; 100 m
A train slows down from 90 km h⁻¹ to 36 km h⁻¹ while covering 150 m. Find the (uniform) acceleration and the time taken.
Given: v₀ = 25 m s⁻¹; v = 10 m s⁻¹; x = 150 m
Use: v² = v₀² + 2ax and v = v₀ + at
a = (v² − v₀²)/(2x) = (100 − 625)/300 = −1.75 m s⁻².
t = (v − v₀)/a = (10 − 25)/(−1.75) = 8.57 s ≈ 8.6 s.
Answer: a = −1.75 m s⁻² (retardation 1.75 m s⁻²); t ≈ 8.6 s
A car moving at 72 km h⁻¹ is braked with a constant deceleration of 5 m s⁻². Find the stopping distance. What would it be at 144 km h⁻¹?
Given: v₀ = 20 m s⁻¹; a = −5 m s⁻²
Use: 0 = v₀² + 2ax
d = v₀²/(2|a|) = 400/10 = 40 m.
At 144 km h⁻¹ = 40 m s⁻¹: d = 1600/10 = 160 m, which is 4 times as large (d ∝ v₀²).
Answer: 40 m; 160 m
A body starts from rest with a constant acceleration of 4 m s⁻². Find the distance it covers in the 5th second.
Given: u = 0; a = 4 m s⁻²
Use: Distance in the 5th second = x(5) − x(4)
x(5) = ½ × 4 × 25 = 50 m; x(4) = ½ × 4 × 16 = 32 m.
Distance in the 5th second = 50 − 32 = 18 m.
(Check with ½ a (2n − 1) = ½ × 4 × 9 = 18 m.)
Answer: 18 m
Car A moves at a constant 20 m s⁻¹ and is 100 m ahead of car B, which starts from rest at that instant with an acceleration of 2 m s⁻². When and where does B catch A, and what is B's speed then?
Given: A: x_A = 100 + 20t; B: x_B = ½ × 2 × t²
Use: Set x_A = x_B
t² = 100 + 20t, so t² − 20t − 100 = 0 and t = 10 + √200 = 24.1 s (the positive root).
Position of B (measured from B's start): t² = 582.8 m, i.e. about 583 m.
Speed of B = 2 × 24.14 = 48.3 m s⁻¹, which is greater than A's speed of 20 m s⁻¹.
Answer: t ≈ 24.1 s; about 583 m from B's start; v_B ≈ 48 m s⁻¹
A bullet moving at 400 m s⁻¹ penetrates 0.20 m into a fixed wooden block and stops. Find its (uniform) retardation and the time taken to stop.
Given: v₀ = 400 m s⁻¹; v = 0; x = 0.20 m
Use: v² = v₀² + 2ax and v = v₀ + at
a = −v₀²/(2x) = −160000/0.4 = −4.0 × 10⁵ m s⁻².
t = v₀/|a| = 400/(4.0 × 10⁵) = 1.0 × 10⁻³ s = 1 ms.
Answer: retardation 4.0 × 10⁵ m s⁻²; t = 1 ms
A particle starts at x₀ = 5 m with velocity −3 m s⁻¹ and a constant acceleration of +2 m s⁻². Find its position and velocity at t = 4 s, and the time and place where it turns back.
Given: x₀ = 5 m; v₀ = −3 m s⁻¹; a = +2 m s⁻²
Use: x = x₀ + v₀t + ½at² and v = v₀ + at
At t = 4 s: x = 5 − 12 + 16 = 9 m; v = −3 + 8 = +5 m s⁻¹.
It turns back when v = 0: −3 + 2t = 0, t = 1.5 s.
Position then: x = 5 − 4.5 + 2.25 = 2.75 m. (It first moves in the negative direction, then returns.)
Answer: x = 9 m, v = +5 m s⁻¹ at 4 s; turns at t = 1.5 s at x = 2.75 m
A car stops in 10 m from 11 m s⁻¹. Assuming the same deceleration, predict the stopping distances from 15, 20 and 25 m s⁻¹.
Given: d₁ = 10 m at v₁ = 11 m s⁻¹
Use: d ∝ v₀²
d = 10 × (v/11)². For 15 m s⁻¹: 10 × 1.86 = 18.6 m.
For 20 m s⁻¹: 10 × 3.31 = 33.1 m. For 25 m s⁻¹: 10 × 5.17 = 51.7 m.
These are close to the measured values 20, 34 and 50 m.
Answer: ≈ 18.6 m, 33.1 m and 51.7 m
A stone is dropped from the top of a 78.4 m high building. Find the time taken to reach the ground and its speed on arrival (g = 9.8 m s⁻²).
Given: u = 0; y = 78.4 m
Use: y = ½gt² and v = gt
t = √(2 × 78.4/9.8) = √16 = 4.0 s.
v = 9.8 × 4.0 = 39.2 m s⁻¹.
Answer: 4.0 s; 39.2 m s⁻¹
A ball is thrown vertically upward at 14.7 m s⁻¹. Find the maximum height, the time to reach it and the total time of flight (g = 9.8 m s⁻²).
Given: v₀ = 14.7 m s⁻¹; at the top v = 0
Use: 0 = v₀² − 2gh and 0 = v₀ − gt
h = v₀²/2g = 216.09/19.6 = 11.0 m (11.025 m).
Time to the top = 14.7/9.8 = 1.5 s.
Time of flight (back at the same level) = 2 × 1.5 = 3.0 s.
Answer: h ≈ 11.0 m; 1.5 s up; 3.0 s in all
A ball is thrown vertically upward at 15 m s⁻¹ from a platform 20 m above the ground (g = 10 m s⁻²). Find the maximum height above the platform, the time to reach the ground and the speed on reaching the ground.
Given: y₀ = 20 m; v₀ = +15 m s⁻¹; a = −10 m s⁻²
Use: Take the ground as y = 0 and use the equation for the whole motion
Height above the platform: h = 15²/(2 × 10) = 11.25 m.
0 = 20 + 15t − 5t², so t² − 3t − 4 = 0, (t − 4)(t + 1) = 0 and t = 4 s (the positive root).
v = 15 − 10 × 4 = −25 m s⁻¹: the speed is 25 m s⁻¹, downward.
Answer: 11.25 m; 4 s; 25 m s⁻¹
A body falls from rest. Find the distances it covers in the 1st, 2nd and 3rd second (g = 9.8 m s⁻²) and show that they are in the ratio 1 : 3 : 5.
Given: u = 0; g = 9.8 m s⁻²
Use: y = ½gt²
After 1 s: 4.9 m; after 2 s: 19.6 m; after 3 s: 44.1 m.
Distances in the 1st, 2nd and 3rd second: 4.9 m, 19.6 − 4.9 = 14.7 m, 44.1 − 19.6 = 24.5 m.
Ratio 4.9 : 14.7 : 24.5 = 1 : 3 : 5.
Answer: 4.9 m, 14.7 m, 24.5 m; ratio 1 : 3 : 5
A ruler dropped between a person's thumb and forefinger falls 18.0 cm before it is caught. Estimate the reaction time (g = 9.8 m s⁻²).
Given: d = 0.180 m; u = 0
Use: d = ½ g t²
t = √(2d/g) = √(0.36/9.8) = √0.03673 = 0.19 s.
Answer: about 0.19 s
A stone is dropped from the top of an 80 m tower at the same instant that another is thrown upward from the foot of the tower at 40 m s⁻¹. When and where do they meet? (g = 10 m s⁻²)
Given: Dropped: y₁ = 80 − 5t²; thrown: y₂ = 40t − 5t²
Use: Set y₁ = y₂ (the same acceleration cancels)
80 − 5t² = 40t − 5t², so 40t = 80 and t = 2 s.
Height above the ground: y = 80 − 5 × 4 = 60 m.
Relative velocity method: each moves with acceleration −g, so relative to the dropped stone the thrown one moves at 40 m s⁻¹ and covers 80 m in 80/40 = 2 s.
Answer: after 2 s, 60 m above the ground
A train A, 150 m long, moving at 72 km h⁻¹, overtakes a train B, 100 m long, moving at 54 km h⁻¹ on a parallel track in the same direction. How long does the overtaking take (from the front of A reaching the rear of B until the rear of A leaves the front of B)?
Given: v_A = 20 m s⁻¹; v_B = 15 m s⁻¹; lengths 150 m and 100 m
Use: Relative velocity: v_AB = v_A − v_B
Relative speed = 20 − 15 = 5 m s⁻¹.
Relative distance to cover = 150 + 100 = 250 m.
Time = 250/5 = 50 s.
Answer: 50 s
Two trains, each 125 m long, move towards each other on parallel tracks at 54 km h⁻¹ and 36 km h⁻¹. How long do they take to cross each other completely?
Given: v₁ = +15 m s⁻¹; v₂ = −10 m s⁻¹
Use: Relative velocity v₁ − v₂ = 15 − (−10)
Relative speed = 15 + 10 = 25 m s⁻¹.
Relative distance = 125 + 125 = 250 m.
Time = 250/25 = 10 s.
Answer: 10 s
A police car moving at 90 km h⁻¹ chases a thief's car moving at 72 km h⁻¹ on a straight road, 100 m ahead. How long does it take to catch the thief?
Given: v_P = 25 m s⁻¹; v_T = 20 m s⁻¹; gap 100 m
Use: Relative speed = 5 m s⁻¹
Relative speed = 25 − 20 = 5 m s⁻¹ (90 − 72 = 18 km h⁻¹ = 5 m s⁻¹).
Time = 100/5 = 20 s.
In that time the police car covers 25 × 20 = 500 m.
Answer: 20 s
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A body can be treated as a point object when its size is much smaller than the distance it moves, and when its own rotation or shape does not matter.
From the graph: the line of A starts at the origin O at t = 0 and ends at the level of P; the line of B starts later (on the time axis, to the right of O), is steeper and ends at the level of Q, which is higher than P.
Measure x from her home along the road.
The graph (scales: time on the horizontal axis, position in km on the vertical axis) is made of three straight segments: from (9:00 am, 0) to (9:30 am, 2.5 km); a horizontal line to (5:00 pm, 2.5 km); and a steep line down to (5:06 pm, 0).
In each cycle of 8 steps (8 s) he moves 5 m forward and 3 m back: a net gain of 2 m per 8 s. The x–t graph is a zig-zag that rises on average.
By calculation: after n complete cycles his position is 2n metres, and in the next cycle he goes 5 m forward before stepping back. He first reaches 13 m during a forward run when 2n + 5 ≥ 13, so n = 4. After 4 cycles (32 s) he is at 8 m. Five more forward steps (5 s) take him to 13 m.
Time taken = 32 s + 5 s = 37 s (he falls into the pit at the 37th step). The graph shows the same: the zig-zag first touches the 13 m line at t = 37 s.
v₀ = 126 km h⁻¹ = 126 × 1000/3600 = 35 m s⁻¹; v = 0; x = 200 m.
From v² = v₀² + 2ax: a = (0 − 35²)/(2 × 200) = −1225/400 = −3.06 m s⁻². The retardation is 3.06 m s⁻².
From v = v₀ + at: t = (0 − 35)/(−3.06) = 11.4 s.
(a) The acceleration is that due to gravity, 9.8 m s⁻², directed vertically downward, during the whole motion (up and down).
(b) At the highest point the velocity is zero and the acceleration is still 9.8 m s⁻² downward.
(c) With the origin at the highest point and downward taken as positive: during the upward motion (t < 0) the ball is below the origin, so the position x is positive; it moves upward, i.e. in the negative direction, so the velocity is negative; the acceleration is downward, so it is positive. During the downward motion (t > 0) the position is positive, the velocity is positive and the acceleration is positive.
(d) Height: v² = u² − 2gh with v = 0: h = 29.4²/(2 × 9.8) = 44.1 m. Time to the top = u/g = 29.4/9.8 = 3 s. The ball returns to the player's hands after 2 × 3 = 6 s.
(a) True. A ball thrown upward has zero speed at the highest point, but its acceleration is g = 9.8 m s⁻² downward at that instant.
(b) False. Speed is the magnitude of velocity. If the speed is zero, the velocity is zero.
(c) True (in one-dimensional motion). Constant speed means the magnitude of the velocity does not change. In one dimension the direction can change only by passing through zero speed (or by an impossible instantaneous reversal), so a particle with a constant non-zero speed has constant velocity and zero acceleration.
(d) False. The particle speeds up only if the acceleration has the same sign as the velocity. A particle moving in the negative direction with positive acceleration slows down. Example: a ball moving upward with the downward direction taken as positive has positive acceleration (+g) and its speed decreases.
Take g = 9.8 m s⁻². Fall: t₁ = √(2 × 90/9.8) = 4.29 s and the speed on arrival is v₁ = 9.8 × 4.29 = 42.0 m s⁻¹.
First collision: the ball loses one tenth of its speed, so it leaves the floor at 0.9 × 42.0 = 37.8 m s⁻¹.
Rise: it rises for 37.8/9.8 = 3.86 s, until t = 4.29 + 3.86 = 8.14 s, when its speed is zero. Second fall: its speed increases again and reaches 37.8 m s⁻¹ at t = 8.14 + 3.86 = 12.0 s, when it hits the floor a second time.
Graph (speed against time): a straight line from 0 at t = 0 up to 42.0 m s⁻¹ at 4.29 s; a sudden drop to 37.8 m s⁻¹; a straight line falling to 0 at 8.14 s; and a straight line rising to 37.8 m s⁻¹ at 12 s.
(a) Displacement is the change of position, x₂ − x₁, measured along the straight line from the initial to the final position. Path length is the total length actually travelled. Example: a man walks 3 m east and then 1 m west. His displacement is 2 m east (magnitude 2 m) but the path length is 4 m. The path length is always at least |displacement| because the shortest route between two points is the straight line. Equality holds when the particle moves in one direction without turning back.
(b) Average velocity = displacement/time, so its magnitude is |Δx|/Δt. Average speed = path length/time. In the example above, over 4 s the magnitude of average velocity is 2/4 = 0.5 m s⁻¹, while the average speed is 4/4 = 1 m s⁻¹. Since path length ≥ |Δx|, the average speed is ≥ the magnitude of the average velocity. Equality holds only for motion in a single direction (no turning back); in that case path length = |displacement|.
Going to the market: time = 2.5/5 = 0.5 h = 30 min. Returning home: time = 2.5/7.5 = 1/3 h = 20 min. So he reaches home at t = 50 min.
(i) 0 to 30 min: displacement 2.5 km in 0.5 h. (a) Magnitude of average velocity = 5 km h⁻¹. (b) Average speed = 5 km h⁻¹.
(ii) 0 to 50 min: displacement 0, path length 5 km, time 50/60 h = 0.833 h. (a) Magnitude of average velocity = 0. (b) Average speed = 5/0.833 = 6 km h⁻¹.
(iii) 0 to 40 min: in the last 10 min he walks back 7.5 × (10/60) = 1.25 km, so he is 2.5 − 1.25 = 1.25 km from home. Displacement = 1.25 km, time = 40/60 = 2/3 h. (a) Magnitude of average velocity = 1.25/(2/3) = 1.875 km h⁻¹. Path length = 2.5 + 1.25 = 3.75 km. (b) Average speed = 3.75/(2/3) = 5.625 km h⁻¹.
Instantaneous velocity is the limit of Δx/Δt as Δt → 0. In such an infinitesimally small interval the particle cannot turn back, so the path length covered equals the magnitude of the displacement: |Δs| = |Δx|.
Therefore the instantaneous speed, lim (Δs/Δt), equals the magnitude of the instantaneous velocity, lim |Δx|/Δt = |dx/dt|. The difference between the two (which arises from turning back during a finite interval) disappears in the limit.
All four graphs cannot represent one-dimensional motion.
No. An x–t graph is a plot of position against time, not the path of the particle. The motion is one-dimensional (along a straight line) at all times.
For t < 0 the graph is flat at x = 0: the particle is at rest at the origin. For t > 0 the graph is a parabola x ∝ t², which means the particle moves along the straight line with constant acceleration.
Suitable context: a ball released from rest at the origin at t = 0 and falling freely (x = ½ g t², measured downward), or a car at rest that starts moving with uniform acceleration at t = 0.
Speed of the van = 30 km h⁻¹ = 8.33 m s⁻¹. Speed of the thief's car = 192 km h⁻¹ = 53.33 m s⁻¹.
Speed of the bullet relative to the ground = 150 + 8.33 = 158.33 m s⁻¹. What damages the car is the speed of the bullet relative to the car:
v = 158.33 − 53.33 = 105 m s⁻¹.
(a) x–t graph: the particle is at rest at x = 0, then moves with constant velocity in the positive direction (rising line to A), then reverses and moves back with constant velocity, passes the origin and goes to negative x (point B), where it stops and stays at rest. Example: a person who is standing at a gate, walks away along a straight lane, returns past the gate to a point on the other side, and stops there.
(b) v–t graph: several straight lines of the same slope, each of smaller height, with sudden jumps between them. Example: a ball bouncing on the floor. Taking upward as positive, the velocity falls uniformly at the rate g, changes sign at the highest point, and at each impact with the floor changes suddenly (reverses with reduced magnitude). Each rise is shorter than the previous one and the bounces come closer in time.
(c) a–t graph: acceleration is zero, then a short pulse, then zero again. Example: a cricket ball (or a hockey puck) moving at constant velocity is struck by a bat for a short time: the force, and hence the acceleration, acts only for a very short interval.
From the graph, x crosses zero going downward at t = 0, and is a sine-like curve with zero crossings at every integer value of t, a trough near t = 0.5 s and a crest near t = 1.5 s. The velocity is the slope of the graph and the acceleration has the sign opposite to the position (a ∝ −x).
The average velocity over an interval is the slope of the chord joining the ends of the interval; it is steepest in interval 3 and gentlest in interval 2.
For motion in one direction in each interval, the average speed equals the magnitude of the average velocity, so the ranking is the same.
The speed–time graph is the |v|–t graph. The average acceleration over an interval is the slope of the chord; the average speed is the average height of the curve.