Question

30. A cyclist is travelling at 15 m s-1. She applies brakes so that she does not collide with a wall 18 m away.
What deceleration must she have ?​

Answer 1

Answer:

In this question we will calculate the negative acceleration/deceleration, as follows:

Kinematic equations:

SUVAT-- s-displacement, u-initial velocity, v-final velocity,a- acceleration, t-timev=u + atv2 = u2 + 2ass=1/2(u+v)ts=ut + 1/2at2

Her initial velocity, u is 15m/s

Her final velocity is, v 0m/s (since she stops)

Displacement is, s 18m

Hence the acceleration, a :

v2=u2 + 2as

0=15^2 + 2 x a x 18

0=225 + 36a

36a= -225

a =  -225/36

a= -6.25 m/s/s

Therefore she has to decelerate at the rate of 6.25m/s2

(or)

u=15m/s

v=0m/s (since she stops)

s=18m

a:

v2=u2+2as

0 = 15^2 + 2 x a x 18

0 = 225 + 36a

36a = -225

a = -225/36

a = -6.25m/s2

Answer 2

Given:-

→Initial velocity of the cyclist=15m/s

→Distance of the cyclist from the wall=18m

To find:-

•Deceleration(negative acceleration)

Solution:-

Since,the cyclist has to finally stop,to avoid collision with the wall,we must take final velocity(v) of the cyclist as 0.

Here,initial velocity,final velocity,and the distance is given.So,to find acceleration:-

By using the 3rd equation of motion,we get:-

=>v²-u²=2as

=>0-(15)²=2×a×18

=>-225=36a

=>a= -225/36

=>a= -6.25m/s².

Since we got acceleration as -6.25m/s²,thus deceleration will be +6.25m/s².

Thus,the cyclist must decelerate at a rate of 6.25m/s² to avoid collision with the wall.