91Ó°ÊÓ

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

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

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

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:

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

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