Question

A sprinter accelerates from rest to 10.0m/s in 1.35s.

Answer 1

Answer:

The average acceleration rate of sprinter who accelerates from rest to 10.0 m/s is 7.407 m/s.

The sprinter can travel 13.5 m in 1.35 s.

Explanation:

From given we came to know that the sprinter can travel at from rest to 10.0 m/s in 1.35 s and we need to calculate average acceleration rate and distance he can travel in 1.35 s.  

To find average acceleration rate, we know that average acceleration rate is ratio of change in velocity to change in time. Represented as follows:

$\text{ Average acceleration rate }=\frac{\text{ Change in velocity }}{\text{ Change in time }}$

We know that change in velocity as 0 m/s to 10.0 m/s and change time as 1.35 s.

$\text{ Average acceleration rate }=\frac{10-0}{1.35}$

$\therefore \text{ Average acceleration rate }=7.407 \ \mathrm{m} / \mathrm{s}^{2}$

Thus, average acceleration rate of sprinter will be $\bold{7.407 \ \mathrm{m} / \mathrm{s}^{2}}$

To find distance he can travel in given interval of time, the velocity is given as:

$\text {Velocity}=\frac{\text {Distance}}{\text {Time}}$

Given that, velocity is 10.0 m/s and time is 1.35 s.

$\Rightarrow 10.0=\frac{\text { Distance }}{1.35}$

$\Rightarrow 10.0 \times 1.35=\text { Distance }$

$\therefore \text { Distance }=13.5 \ \mathrm{m}$

Therefore, sprinter can travel a distance of 13.5 m.

Answer 2

Answer:

a) Acceleration = 7.41 m/s^2

b) Distance ran = 6.75 m

Explanation:

For Acceleration use the formula...

a = (v - u) ÷ t....where v=10 and u=0....and t=1.35

a = (10 - 0)÷ 1.35

;acceleration = 7.41m/s^2

For the distance moved in 1.35s

;use the...s = ut + 1/2at^2....where s is the distance moved

;s = (0)(1.35) + 0.5(7.41 × 1.35^2)

Distance moved = 6.75m