Question

(9) A car travelling at a speed of 10 m/s stops in 4 sec after the application of brakes. If we assume the
retardation to be constant, then
a) Draw a speed-time graph for the above information.
b) Find the retardation,
c) Find the distance travelled during this time.​

Answer 1

Answer:

Explanation:

V=0m/sec

U=10 m/sec

a=-2 m/sec*2

Now,

V*2 - U*2 =2aS

0–100= - 2*2S

S = 25m

Answer 2

Dear Friend,

◆ Answers -

a = -2.5 m/s^2

s = 20 m

◆ Explanation -

# Given -

u = 10 m/s

v = 0

t = 4 s

# Solution -

(i) Refer to image for speed-time graph.

(ii) Retardation -

Newton's 1st law of Kinetics -

v = u + at

a = (v - u) / t

a = 0 - 10 / 4

a = -2.5 m/s^2

(iii) Stopping distance -

Newton's 1st law of Kinetics -

s = ut + 1/2 at^2

s = 10×4 + 1/2 × (-2.5)×4^2

s = 40 - 20

s = 20 m

Hope this helps you.