(a) Explain the terms:- uniform acceleration and average speed.
(b) A body at rest is given an initial uniform acceleration of 8.0ms\(^{-2}\) for 30s after which the acceleration is reduced to 5.0ms\(^{-2}\) for the next 20s. The body maintains the speed attained for 60s after which it is brought to rest in 20s. Draw the velocity-time graph of the motion using the information given above.
(c) Using the graph, calculate the: (i) maximum speed attained during the motion; (ii) average retardation as the body is being brought to rest; (iii) total distance travelled during the first 50s; (iv) average speed during the same interval as in (ii).
(b) A body at rest is given an initial uniform acceleration of 8.0ms\(^{-2}\) for 30s after which the acceleration is reduced to 5.0ms\(^{-2}\) for the next 20s. The body maintains the speed attained for 60s after which it is brought to rest in 20s. Draw the velocity-time graph of the motion using the information given above.
(c) Using the graph, calculate the: (i) maximum speed attained during the motion; (ii) average retardation as the body is being brought to rest; (iii) total distance travelled during the first 50s; (iv) average speed during the same interval as in (ii).
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Correct Answer: Option n
Explanation:
(a) Uniform acceleration occurs when the change in velocity per unit time is constant. The S.I. unit is ms\(^{-2}\).
\(a = \frac{v - u}{t}\); a - uniform acceleration, v - final velocity, u - initial velocity, t - time.
Graphically,
acceleration is the same acceleration at x is different every point from that at Y.
Note - acceleration is given by slope in both graphs.
(b)
(i) Maximum speed = 240 m/s.
(ii) \(a = \frac{v - u}{t} = \frac{0 - 200}{20} = -10 ms^{-2}\)
(iii) Area of I = \(\frac{1}{2} \times 30 \times 240 = 3600m\)
Area of II = \(20 \times 200 = 4000m\)
Area of III = \(\frac{1}{2} \times 20 \times 40 = 400m\)
Total distance in the first 50s : \(3600m + 4000m + 400m = 8000m\)
(iv) Average speed during the first 50s = \(\frac{8000m}{50s} = 160 ms^{-1}\)
(a) Uniform acceleration occurs when the change in velocity per unit time is constant. The S.I. unit is ms\(^{-2}\).
\(a = \frac{v - u}{t}\); a - uniform acceleration, v - final velocity, u - initial velocity, t - time.
Graphically,
acceleration is the same acceleration at x is different every point from that at Y.
Note - acceleration is given by slope in both graphs.
(b)
(i) Maximum speed = 240 m/s.
(ii) \(a = \frac{v - u}{t} = \frac{0 - 200}{20} = -10 ms^{-2}\)
(iii) Area of I = \(\frac{1}{2} \times 30 \times 240 = 3600m\)
Area of II = \(20 \times 200 = 4000m\)
Area of III = \(\frac{1}{2} \times 20 \times 40 = 400m\)
Total distance in the first 50s : \(3600m + 4000m + 400m = 8000m\)
(iv) Average speed during the first 50s = \(\frac{8000m}{50s} = 160 ms^{-1}\)