From rest, the sprinter reach his stop speed of 15 m/s at a constant acceleration.

So we need to determine first the distance covered by the sprinter when he is running at a constant acceleration. Let's use the formula:

`v_2^2=v_1^2+2ad`

where v1 - initial velocity

...

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

From rest, the sprinter reach his stop speed of 15 m/s at a constant acceleration.

So we need to determine first the distance covered by the sprinter when he is running at a constant acceleration. Let's use the formula:

`v_2^2=v_1^2+2ad`

where v1 - initial velocity

v2 - final velocity

d - distance

a - acceleration and

t - time

Then, substitute `v_1=0` , `v_2=15m//s` and `a=2.5m//s^2` . Also, let `d_a` the distance covered by the sprinter at constant acceleration.

`15^2=0^2+2(2.5)d_1`

`225= 5d_1`

`d_a=45`

So, `d_a=45 m` .

Moreover, we still need to determine the time it takes the sprinter to reach his maximum speed. Use the formula:

`v_2=v_1+at`

Substitute `v_1=0` , `v_2=15` and `a=2.5` . Also, let `t_a` the time it takes the sprinter to reach his top speed at constant acceleration.

`15=0+2.5t_a`

`15=2.5t_a`

`t_a=6`

Hence, `t_a=6s` .

When the sprinter reaches his top speed, he continues to run at a speed of 15 m/s. This indicates that the sprinter is no longer running at constant acceleration. Instead he is running at a constant velocity.

To solve for the total time it takes for him to finish the 100-m dash, we need to determine the remaining distance he need still need to run. Let the remaining distance be `d_b` .

`d_a+d_b=100`

`45+d_b=100`

`d_b=55`

So, `d_b=55 m` .

Next, let's determine the time it takes the sprinter to run the 55m at constant velocity. Let's use the formula:

`d=vt`

Substitute `d_a=55m` and `v=15m//s` . Let time be `t_b` .

`55=15t_b`

`t_b=55/15`

`t_b=11/3`

So the total time it takes the sprinter to finish the 100-m dash is:

`t=t_a+t_b`

`t=6+11/3`

`t=29/3=9.67`

--------------------------------------------------------------------------------------- **Hence, it takes 9.67s for the sprinter to run the 100-m dash.**