Question

a truck starts from rest with a acceleration of 1.5 ms-2 while a car 150 m behind the truck starts from rest with an acceleration of 2 ms-2. how long will it take before both the truck and car side by side.

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{distance}\mathrm{covered}\mathrm{by}\mathrm{truck}\mathrm{in}\mathrm{time}\mathrm{t}\mathrm{is}\mathrm{given}\mathrm{by}, \\ \mathrm{s}=\frac{1}{2}\times 1.5\times {\mathrm{t}}^2---\left(1\right) \\ \mathrm{Similarly}\mathrm{the}\mathrm{distance}\mathrm{covered}\mathrm{by}\mathrm{the}\mathrm{car}\mathrm{in}\mathrm{time}\mathrm{t}, \\ \mathrm{s}+150=\frac{1}{2}\times 2\times {\mathrm{t}}^2---\left(2\right) \\ \mathrm{from}\mathrm{eqns}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get}, \\ \frac{1}{2}\times 1.5\times {\mathrm{t}}^2+150=\frac{1}{2}\times 2\times {\mathrm{t}}^2 \\ \frac{1}{2}\times (2-1.5)\times {\mathrm{t}}^2=150. \\ \frac{1}{2}\times 0.5\times {\mathrm{t}}^2=150 \\ {\mathrm{t}}^2=\frac{2\times 150\times 10}{5}=600 \\ \mathrm{t}=10\sqrt{6}\mathrm{s}=10\times 2.45=24.5\mathrm{s}. \\ \mathrm{Hence}\mathrm{the}\mathrm{car}\mathrm{will}\mathrm{catch}\mathrm{the}\mathrm{truck}\mathrm{in}24.5\mathrm{second}\text{'}\mathrm{s}\mathrm{time} \end{gathered}$$