Question

An object moving with 6 m/sec executes an acceleration of 2 m/s² in the next 3 seconds. How much distance did it cover?

Answer 1

Step (1) Given:

The initial velocity of the object, $u=6m/s$

The acceleration of the object, $a=2m/s^2$

Time for which the object moved, $t=3sec$

Step (2) Finding the distance covered by the object using the equation of motion:

We know that,

$$\begin{gathered} Dis\tan ce\left(s\right)=ut+\frac{1}{2}a{t}^2 \\ Bysubstitutingthegivenvaluesintheexpression,weget \\ s=6\times 3+\frac{1}{2}\left(2\right){\left(3\right)}^2=18+\frac{1}{2}\left(2\right)\left(9\right) \\ =18+\left(9\right) \\ =27m \end{gathered}$$

Therefore, the object covered 27 m.