• question_answer A car accelerates from rest in the 1st part of its journey with a m/s2 and then it retards with 2a m/s2 so as to bring it to rest again. If t is the total time of journey, then the distance travelled by car is A)  $\frac{a{{t}^{2}}}{3}$                  B)  $\frac{a{{t}^{2}}}{2}$ C)  $a{{t}^{2}}$                    D)  $\frac{a{{t}^{2}}}{4}$

Idea It is a simple question based on kinematics equations. First apply the equation for acceleration and then for retardation. Let, ${{t}_{0}}$is the time during acceleration then$\frac{{{t}_{0}}}{2}$will be the time of de-acceleration in order to bring car to rest again. So, time of journey$t={{t}_{0}}+\frac{{{t}_{0}}}{2}=\frac{3{{t}_{0}}}{2}$ $\therefore$${{t}_{0}}=\frac{2t}{3}$ and$\frac{{{t}_{0}}}{2}=\frac{t}{3}$ Hence, $s=\frac{1}{2}a\times {{({{t}_{0}})}^{2}}+\frac{1}{2}\times 2a\times {{\left( \frac{{{t}_{0}}}{2} \right)}^{2}}$ $=\frac{1}{2}a{{\left( \frac{2t}{3} \right)}^{2}}+\frac{1}{2}2a\times {{\left( \frac{t}{3} \right)}^{2}}=\frac{a{{t}^{2}}}{3}$ TEST Edge In kinematics different types of questions could be asked, e.g., if a body is in free fall draw the graph between${{v}^{2}}$and s (reference point could be considered at any point). Let us consider origin at the highest point from where we have left the object. For free fall, ${{v}^{2}}={{u}^{2}}+2as$ $\therefore$$u=0$ ${{v}^{2}}=2as$ ${{v}^{2}}/s=2a=$constant You will be redirected in 3 sec 