An antelope moving with constant acceleration covers the distance between two points $70.0$apart in$6.00s$. Its speed as it passes the second point is$15.0m/s$. What are (a) its speed at the first point and (b) its acceleration?

• The distance between the two points is $70.0m$.
• The distance between the two points has been covered in$6.00s$ .

This law describes that without the influence of an external force, a body will continue to move along a straight line at a uniform speed.

The speed of the antelope can be identified with the help of averaging the initial and the final velocity and multiplying it by the time. Moreover, the acceleration of the antelope can also be gathered by subtracting the final velocity from the initial velocity and dividing the whole subtraction by time.

a) From Newton’s first law, the displacement of the antelope can be expressed as:

$\Delta x=\left(\frac{v+v_0}{2}\right)t$

Here, $\Delta x$is the distance between the two points, v is the speed of the antelope passing through the second point and t is the time taken to cover the distance between the two points.

Substituting the values in the above expression, we get-

$\begin{array}{rcl}70\text{m}& =& \left(\frac{15m/s+v_{0}m/s}{2}\right)\times 6s\\ 70\text{m}& =& 45\text{m}+3v_0\text{m}\\ v_0& =& 8.34m/s\end{array}$

Thus, the speed at the first point of the antelope is$8.34m/s.$

b) From Newton’s first law, the velocity of the antelope can be expressed as:

$v=v_0+at$

Here, v is the final and $v_o$is the initial velocity of the antelope. Furthermore, a is the acceleration and t is the time taken for covering the distance between the two points.

Substituting the values in the above expression, we get-

$\begin{array}{rcl}15m/s& =& 8.34m/s+a\times 6s\\ a& =& 1.11m/s^2\\ & & \end{array}$

Thus, the acceleration of the antelope is$1.11m/s^2$ .