Question

A sprinter has maximum speed of 10 m/s and reaches that speed by an acceleration of 2.5 m/s^2.The time he takes to run 100 m is

Answer 1

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

Answer 2

Given:

Maximum speed $\[ = 10\,m/s\]$

Acceleration $\[ = 2.5\,m/{s^2}\]$

Distance $\[ = 100\,m\]$

To Find: Time taken to run 100 m

Solution:

According to the first equation of motion, we have,

$\[v = u + at\]$

Therefore, the time required to reach the maximum speed is given as:

$\[\begin{gathered} t = \frac{{v - u}}{a} \hfill \\ t = \frac{{(10 - 0)m/s}}{{2.5\,m/{s^2}}} \hfill \\ t = 4\,s \hfill \\ \end{gathered} \]$

From the second equation of motion, we have,

$\[s = ut + \frac{1}{2}a{t^2}\]$

Therefore, the distance covered in 4 seconds is given by

$\[\begin{gathered} s = 0 \times 4 + \frac{1}{2} \times 2.5 \times {4^2} \hfill \\ s = 2.5 \times 8 = 20\,m \hfill \\ \end{gathered} \]$

Now, time taken to complete the remaining distance, i.e., $100-20=80m$ is given as:

$\[\begin{gathered} t = \frac{{80\,m}}{{10\,m/s}} \hfill \\ t = 8\,s \hfill \\ \end{gathered} \]$

Thus, the total time required to complete 100 m is

Total time $\[ = 4\,s + 8\,s = 12\,s\]$

Hence, the time the sprinter takes to run 100 m is $\[12\,s\]$.

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