Question

How far (in meters) does a person travel in coming to a complete stop in 40 ms at a constant acceleration of 60g?

Answer 1

Answer:

d= 0.48 m

Explanation:

Given that

time ,t= 40 m s= 40 x 10⁻³ s

Constant acceleration ,a= - 60 g

We know that g= 10 m/s²

Therefore ,a=- 60 x 10 =- 600 m/s²

We know that

v= u + a t

u=initial speed

v=final sped

t=time

a=acceleration

Now by putting the value

0 = u - 600 x  40 x 10⁻³

u = 24  m/s

The distance cover before coming to stop d is given as

$d=ut+\dfrac{1}{2}at^2$

$d=24\times 40\times 10^{-3}-\dfrac{1}{2}\times 600\times (40\times 10^{-3})^2$

d= 0.48 m