Question

On entering a state highway, a car continues to move with a constant velocity of 6 m s^–1 for 2 minutes and then accelerates with a constant acceleration 1 m s^–2 for 6 seconds. Find the displacement of the car on the state highway in the 2 min 6 s time interval by drawing a velocity-time graph for its motion.

Answer 1

Given:

Initial velocity, u = 6 m s⁻¹

The car travels at a constant velocity for 2 min = 120 s

Acceleration, a = 1 m s⁻²

Duration of acceleration = 6 s

Final velocity after accelerating:

v = u + at

= 6 + 1 × 6

= 12 m s^–1

The velocity-time graph is shown below:

Displacement = area under the velocity-time graph

The area under the graph has two parts:

Area A (rectangle OPQS):

S₁ = length × breadth

= 120 × 6

= 720 m

Area B (trapezium QRTS):

S₂ = `1/2 xx "sum pf parallel sides" xx "height" = 1/2 xx (6 + 12) xx 6 = 1/2 xx 8 xx 6 = 54` m

Total displacement,

S = S₁ + S₂ = 720 + 54 = 774 m

Hence, the displacement of the car in 2 min 6 s is 774 m.