Question

A car is travelling with a speed of 36km/h. The driver applies the brake and retards the car uniformly. The car is stopped in 5s. Find(1) retardation of the car. (2) distance travelled before it is stopped after applying the brakes.

Answer 1

Given:
u=36km/h=10m/s
t=5s
v=0m/s

1). a=(v-u)/t
a=(0-10)/5
a=-10/5=-2m/s²
so,retardation is 2m/s².

2). v²=u²+2aS
0²=10²+2*(-2)*S
0=100-4*S
S=-100/-4=25m Ans.

Answer 2

Answer:

Retardation = 2 m/s²

Distance traveled = 25 m

Step-by-step explanation :

Given,

u = 36 km/h

= ( 36 *5/18 ) m/s = 10m/s

v = 0

t = 5sec

( i ) retardation of the car

We know,

v = u + at

Substituting the values,

0 = 10 + a*5

0 = 10 + 5a

5a = -10

a = -10 / 5

a= -2 m/s²

Acceleration = -2 m/s²

Since acceleration is negative,

retardation = 2 m/s²

( ii )  distance travelled before it is stopped after applying the brakes

We know,

S = ut + 1/2 at²

Substituting the values,

S = 10*5 + 1/2 * (-2) * (5)²

S = 50 + 1/2* (-2) * 25

S = 50 + 1/2* (-50)

S = 50 + (-25)

S = 25m

So, Distance traveled = 25 m

Hence,

Retardation = 2 m/s²

Distance traveled = 25 m