Question

(a) Derive the formula : $s=ut+\frac{1}{2}a{t}^2$, where the symbols have usual meanings.
(b) A train starting from stationary position and moving with uniform acceleration attains a speed of 36 km per hour in 10 minutes. Find its acceleration.

Answer 1

(a) Consider a body in motion starts with an initial velocity (u) and uniform acceleration (a) and achieves a final velocity (v) in time (t) covering a distance 's'. Then distance 's' traveled by a moving body can be calculated by considering its average velocity for time (t). Since, the average velocity is sum average of initial velocity (u) and final velocity is (v) of the body, i.e.
$$\begin{gathered} \mathrm{Average}\mathrm{velocity}=\frac{\mathrm{Initial}\mathrm{velocity}+\mathrm{Final}\mathrm{velocity}}{2} \\ \mathrm{So},\mathrm{Average}\mathrm{velocity}=\frac{\mathit{v}\mathit{+}\mathit{u}}{\mathit{2}} \end{gathered}$$
Also we know that,
Distance travelled = (Average velocity) (Time)
Using this relationship in equation (1),
$s=\left(\frac{u+v}{2}\right)t$
Now using the 1^st equation of motion v = u + at in the above,
$s=\left(\frac{2u+at}{2}\right)t$
On further simplification we get the 2^nd equation of motion as,
$\boxed{s=ut+\frac{1}{2}a{t}^2}$
Where,
(s) - Displacement
(u) - Initial velocity
(a) - Acceleration
(t) - Time

(b) We have to find the value of uniform acceleration. So,
Initial velocity (u) = 0 m/s
Final velocity
$$\begin{gathered} \left(v\right)=36\mathrm{km}/\mathrm{hr} \\ =\frac{36\left(1000\right)}{3600}\mathrm{m}/\mathrm{s} \\ =10\mathrm{m}/\mathrm{s} \\ \mathrm{Time}\mathrm{taken}, \\ \left(t\right)=\left(10\right)\left(60\right)\mathrm{s} \\ =600 \end{gathered}$$
So acceleration,
$a=\frac{v-u}{t}$
Put the values in the above equation to get the value of acceleration
$$\begin{gathered} a=\frac{10-0}{600} \\ =\boxed{0.0167\mathrm{m}/{\mathrm{s}}^2} \end{gathered}$$