Question

A particle starts from rest and moves along a straight line under the action of a constant force. If it travels a distance $x$ in the first 5 s, what will be the distance travelled by it in the next 5 s?

Answer 1

From the second law of motion,
$F=ma\Rightarrow a=\frac{F}{m}$
If a constant force acts on a body, the acceleration will be constant.

From t = 0 to t = 5 s, applying the second equation of motion,
$x=ut+\frac{1}{2}a{t}^2=\frac{1}{2}\left(\frac{F}{m}\right)5^2=\frac{25}{2}\left(\frac{F}{m}\right)$

Now, from t = 0 to t = 10 s, applying the second equation of motion,
$x^{\prime }=ut+\frac{1}{2}a{t}^2=\frac{1}{2}\left(\frac{F}{m}\right){10}^2=\frac{100}{2}\left(\frac{F}{m}\right)$

So, distance moved from t = 5 s to t = 10 s is given by $x^{\prime }-x$
$=\frac{100}{2}\left(\frac{F}{m}\right)-\frac{25}{2}\left(\frac{F}{m}\right)$
$=\frac{75}{2}\left(\frac{F}{m}\right)=3x$