If a car at rest accelerates uniformly to a speed of $$144\,km/h$$   in $$20\,s,$$  it covers a distance of

Solution :
Initial velocity of car $$\left( u \right) = 0$$
Final velocity of car $$\left( v \right) = 144\,km/hr = 40\,m/s$$
Time taken $$ = 20\,s$$
We know that, $$v = u + at$$
$$\eqalign{ & 40 = a \times 20 \Rightarrow a = 2\,m/{s^2} \cr & {\text{Also,}}\,\,{v^2} - {u^2} = 2as \Rightarrow s = \frac{{{v^2} - {u^2}}}{{2a}} \cr & \Rightarrow s = \frac{{{{\left( {40} \right)}^2} - {{\left( 0 \right)}^2}}}{{2 \times 2}} = \frac{{1600}}{4} = 400\,m. \cr} $$