Question

A vehicle, starting with a velocity of 60 m/s moves with an acceleration of $5\frac{\mathrm{m}}{{\mathrm{s}}^2}$​ . What will be the distance traveled during 10 s.

Answer 1

Step 1:- Given Data

The initial velocity of the vehicle $\mathrm{u}=60\mathrm{m}{\mathrm{s}}^{-1}$

Acceleration of vehicle $\mathrm{a}=5\mathrm{m}{\mathrm{s}}^{-2}$

Time taken by the vehicle $\mathrm{t}=10\sec$

distance s =?

Step 2:- Formula used

Using the second equation of motion

$\mathrm{s}=\mathrm{u}\times \mathrm{t}+\frac{1}{2}\times \mathrm{a}\times {\mathrm{t}}^2$

Step 3:-Calculation for the distance

The distance traveled during 10 sec

$$\begin{gathered} \mathrm{s}=\mathrm{u}\times \mathrm{t}+\frac{1}{2}\times \mathrm{a}\times {\mathrm{t}}^2 \\ \mathrm{s}=60m{s}^{-1}\times 10sec+\frac{1}{2}\times 5m{s}^{-2}\times {\left(10sec\right)}^2 \\ \mathrm{s}=600+250 \\ \mathrm{s}=850\mathrm{m} \end{gathered}$$

Hence the distance covered by the object is 850 m.