Question

The driver of a car A going at 25 ms–1 applies the brakes, decelerates uniformly, and stops in 10s. The driver
of another car B going at 15 ms–1 puts less pressure on this brakes and stops in 20 s. On the same graph,
plot speed-time for each of the two cars.
(a) Which of the two cars travelled farther after the brakes were applied ?
(b) Add a line to the graph, which shows the car B decelerating at the same rate as the car A. How long
does it take the car B to stop at this rate of deceleration ?

Answer 1

Answer:

Explanation:

a) We know that area under v-t graph represent displacement

Disp. of car A = 1/2* base * height

( 1/2)(10)(25) = 125m

Disp. of car B = (1/2)(20)(15) = 150 m

→ So, car B will travel 25 m more after brakes were applied

b) Deceleration of car A can be find out using

first equation of motion

〔 v= u+ at 〕

0 = 25 + a(10)

a = - 2.5 ms-²

Now if car B decelerates with "a"

Time take can be find out by first equation of motion

v = u + at

0 = 15 - 2.5t

t = (-15)/(-2.5) = 6s

Remember one thing 一＞ area under v-t graph represent displacement