Question

A sprinter leaves the starting blocks with an acceleration of 4.5m/s^2. What is the sprinter's speed 2 s later

Answer 1

v=v0+at
v=0 m/s + 4.5 m/s^2 * 2 s
v=9 m/s

Answer 2

9 m/s

Step-by-step explanation:

For a uniformly accelerated motion, the speed at time t is given by:

`v(t) = u +at`

where

u is the initial speed

a is the acceleration

t is the time

In this problem, we have:

u = 0 m/s is the initial speed

a = 4.5 m/s^2 is the acceleration

t = 2 s is the time

Substituting into the equation, we find

`v(2 s)=u+at=0 +(4.5 m/s^2)(2 s)=9 m/s`