Question

a cyclist is travelling at 15m/s and applies brakes so that to not to collide on a wall at 18 m away what deaccelaration must he have?FAST PLEASE!!!​

Answer 1

Answer:

Explanation:

v=0

u=15m/s

s=18m

a=?

V^2 - u^2 = 2as

0 - 225 = 2 × a × 18

-225/36 = a

a = -6.81m/s

Therefore,

Retardation is 6.81m/s

Answer 2

Answer:-

a = -6. 25 m/s²

Given :-

u = 15 m/s

s = 18 m

v = 0 m/s

To find :-

Deceleration produced in cycle.

Solution:-

When the cyclist applies the brakes the cycle retards uniformly and finally comes to rest.

By using third equation of motion:-

$\huge \boxed{2as = v^2 -u^2 }$

Put the given values,

→$2 \times a \times 18 = (0) ^2 -(15) ^2$

→$36 a = -225$

→$a = \dfrac{-225}{36}$

→$a = -6. 25 m/s^2$

hence,

The required retardation will be -6. 25 m/s².