Derive third equation of motion by graphical method.

Derive the third equation of motion by a method other than the graphical method.

A train starting from rest attains a velocity of $72 \mathrm{km} {\mathrm{h}}^{-1}$ in $5$ minutes. Assuming that the acceleration is uniform. Find

(i) the acceleration

(ii) the distance travelled by the train for attaining this velocity.

A car accelerates uniformly from $18 \mathrm{km} {\mathrm{h}}^{-1}$ to $36 \mathrm{km} {\mathrm{h}}^{-1 }$in $5 \mathrm{s}$. Calculate
(i) the acceleration 
(ii) the distance covered by the car in that time.

A car moving along a straight line at a speed of $72 \mathrm{km}$ ${\mathrm{h}}^{-1}$stops in $5 \mathrm{s}$ after the brakes are applied.

(a) Find the acceleration

(b) Plot the graph of speed versus time

(c) Using the graph find the distance covered by the car after the brakes are applied.

An object is dropped from a height of $120 \mathrm{m}$. The acceleration of the object is $10 {\mathrm{ms}}^{-2}$ downwards. Find the distance travelled by the object after 

(a) $1 \mathrm{s}$

(b) $2 \mathrm{s}$

(c) $3 \mathrm{s}$

What is the final velocity of the object after $3 \mathrm{s}$?

An object is moving with a velocity of $+6 {\mathrm{ms}}^{-1}$and with an acceleration of$-1 {\mathrm{ms}}^{-2}$. What will be the
(a) distance travelled by the car before coming to rest?
(b) time taken for coming to rest?