Question

If the sprinter from the previous problem accelerates at that rate for 20 m, and then maintains that velocity for the remainder of the 100-m dash, what will be his time for the race?

Answer 1

Question:

A 63.0 kg sprinter starts a race with an acceleration of 4.20m/s square. What is the net external force on him? If the sprinter from the previous problem accelerates at that rate for 20m, and then maintains that velocity for the remainder for the 100-m dash, what will be his time for the race?

Answer:

Time for the race will be t = 9.26 s

Explanation:

Given data:

As the sprinter starts the race so initial velocity = v₁ = 0

Distance = s₁ = 20 m

Acceleration = a = 4.20 ms⁻²

Distance = s₂ = 100 m

We first need to find the final velocity (v₂) of sprinter at the end of the first 20 meters.

Using 3rd equation of motion

(v₂)² - (v₁)² = 2as₁ = 2(4.2)(20)

v₂ = 12.96 ms⁻¹

Time for 20 m distance = t₁ = (v₂ - v ₁)/a

t₁ = 12.96/4.2 = 3.09 s

He ran the rest of the race at this velocity (12.96 m/s). Since has had already covered 20 meters, he has to cover 80 meters more to complete the 100 meter dash. So the time required to cover the 80 meters will be

Time for 100 m distance = t₂ = s₂/v₂

t₂ = 80/12.96 = 6.17 s

Total time = T = t₁ + t₂ = 3.09 + 6.17 = 9.26 s

T = 9.26 s

Answer 2

The time for the race will be 9.26 sec. The total time for the race is the sum of the time taken in the two parts.

What is acceleration?

The rate of change of the velocity is known as the acceleration. It is a time-based quantity having unit m/sec².

The given data in the problem is;

u is the initial velocity= 0

s₁ is the distance  = 20 m

a is the acceleration= 4.20 ms⁻²

s₂ is the distance 2 =100 m

The final velocity is found by;

$\rm v^2-u^2=2as \\\\ v = \sqrt{2as} \\\\ v = \sqrt{2\times 4.20\times 20} \\\\ v = 12.96 \m/sec$

The time period is found by;

$\rm t_1= \frac{v-u}{a} \\\\ \rm t_1= \frac{12.96-0}{4.2} \\\\ \rm t_1= 3.09 \ sec$²

The time for the remaining 100 m will be;

$\rm t_2= \frac{v}{a} \\\\ \rm t_2= \frac{80}{12.96} \\\\ \rm t_2=6.17 sec$

The total time will be;

$\rm t = t_1+t_2 \\\\ \rm t = 3.09+6.17 \\\\ \rm t =9.26 sec$

Hence time for the race will be 9.26 sec.

To learn more about acceleration refer to the link;

https://brainly.com/question/12134554