Question

A CAR IS TRAVELLING AT A SPEED OF 10 ms. IT ACCELERAGS AT A RATE OF 1 ms FOR 40 SECS. THEN IT MOVES AT A CONSTANT SPEED FOR NEXT 2 MIN. FINALLY THE CAR COMES TO A STOP IN NEXT 90 SECS WITH A CONSTANT RETARDATION. WHAT IS THE TOTAL DISTANCE TRAVELLED BY THE CAR ?

Answer 1

$$\begin{gathered} \mathrm{Case}\mathrm{I}: \\ \mathrm{Final}\mathrm{velocity}: \\ \mathrm{v}=\mathrm{u}+\mathrm{at} \\ =10+\left(1\times 40\right) \\ =50\mathrm{m}/\mathrm{s} \\ \mathrm{Distance}: \\ {\mathrm{s}}_1=\left(\frac{\mathrm{u}+\mathrm{v}}{2}\right)\mathrm{t} \\ =\left(\frac{10+50}{2}\right)40 \\ =1200\mathrm{m} \\ \mathrm{Case}\mathrm{II}: \\ \mathrm{Distance}: \\ {\mathrm{s}}_2=\mathrm{vt} \\ =50\times 120(2\min =120\mathrm{s}) \\ =6000\mathrm{m} \\ \mathrm{Case}\mathrm{III}: \\ \mathrm{Distance}: \\ {\mathrm{s}}_3=\left(\frac{\mathrm{v}+\mathrm{v}\text{'}}{2}\right)\mathrm{t} \\ =\left(\frac{50+0}{2}\right)90 \\ =2250\mathrm{m} \\ \mathrm{Total}\mathrm{distance}: \\ \mathrm{d}=1200+6000+2250 \\ \mathbf{}\mathbf{}\mathbf{=}\mathbf{9450}\mathbf{}\mathbf{m} \end{gathered}$$