A car travels with a uniform velocity of 20 ms⁻¹ for 5 s. The brakes are then applied and the car is uniformly retarded. It comes to rest in further 8 s. Draw a graph of velocity against time. Use this graph to find :
(i) the distance travelled in first 5 s.
(ii) the distance travelled after the brakes are applied.
(iii) total distance travelled, and
(iv) acceleration during the first 5 s and last 8 s.
Answers
i) The distance travelled by the car in the first 5 sec is 100 m.
ii) The distance travelled by the car in the last 8 sec is 80 m.
iii) Total distance travelled by the car is 180 m.
iv) Acceleration during the first 5 sec is 0 ms⁻², and retardation during the last 8 sec is 2.5 ms⁻².
Refer to the graph attached below for all calculations.
i) Given,
Initial velocity (u) = 20 ms⁻¹
Time (t) = 5 sec
Therefore, distance travelled in the first 5 sec (S) = Rectangular area under the velocity-time graph from 0 sec to 5 sec.
=> S = length × breadth
=> S = 5 s × 20 ms⁻¹
=> S = 100 m
ii) Time after which the brakes are applied (t') = 8 sec
The car comes to rest,
=> Final velocity (v) = 0
• Therefore, the distance travelled (S') after the brakes are applied = Triangular area enclosed by the velocity-time graph from 5 sec to 13 sec
• Therefore, distance (S') = (1/2) × 8 sec × 20 ms⁻¹
=> S' = 80 m
iii) Total distance travelled = S + S'
=> Total distance= 100 m + 80 m
=> Total distance = 180 m
iv) • From the graph, acceleration during the first 5 sec = slope of the graph
• Slope of a graph = perpendicular / base
• Since, the graph for the first 5 sec is a rectangle, the slope is zero.
• Therefore, the acceleration for the first 5 sec is 0, since the car moves at a constant velocity.
• Acceleration during the last 8 seconds = Slope of the triangular graph
=> slope = perpendicular (p) / base (b)
• p = 20 ms⁻¹, b = (5 - 13) sec = -8 sec
Therefore, slope = 20 ms⁻¹ / (-8) sec
=> slope = -2.5 ms⁻²
Therefore, acceleration = -2.5 ms⁻², i.e., retardation = 2.5 ms⁻².
