Question

A sprinter can accelerate with constant acceleration for 4.0 s before reaching top speed. He can run the 100 meter dash in 10.0 s. What is his speed as he crosses the finish line?

Short answer

The speed of the sprinter when he crosses the finish line is 12.5 m/s

Step-by-step solution

Draw the diagram for the given problem:

In the above diagram, assume that the motion is along the positive X-axis.

Divide the motion into two parts:

1. From t=0s to t= 4s

2. From t=4s to t=10s

Calculating the maximum velocity at t=4s and position at 4s:

Since the person is moving with constant acceleration from t=0s to 4s, apply the kinematic equation of motion to calculate the maximum velocity, v_1x.

$$\begin{gathered} v_{1x}=v_{0x}+a_{0x}{t}_1 \\ =0m/s+4a_{0x} \\ =4a_{0x} \end{gathered}$$

In the above equation, v_0x is the initial velocity of the person at t=0s.

The formula to calculate the position of the person at t=4s is given by

$$\begin{gathered} x_1=x_0+v_{0x}{t}_1+\frac{1}{2}{a}_{0x}{t_1}^2 \\ =0+0+\frac{1}{2}{a}_{0x}{4}^2 \\ =8a_{0x} \end{gathered}$$

Calculating the final velocity as the person crosses the finish line:

The length of the dash board is given as 100m.

The kinematic equations of motion to calculate the acceleration is given by :

$$\begin{gathered} x_2=x_1+v_{1x}(t_2-t_1) \\ 100=8a_{0x}+4a_{0x}(10-4) \\ 100=32a_{0x} \\ {a}_{0x}=\frac{100}{32} \\ =3.125m/s^2 \end{gathered}$$

The velocity of the person at the end of the cross line is taken as v_2x and it is equal to v_1x.

Therefore, solving the above equation and substituting the value of a_0x:

$$\begin{gathered} v_{2x}=v_{1x}=4a_{0x} \\ =4\times 3.125 \\ =12.5m/s \end{gathered}$$

Therefore, the speed of the sprinter as he crosses the finish line is 12.5 m/s.