A body which is uniformly accelerated, changes its velocity from $36 k m / h r$ in one direction to $18 k m / h r$ in the opposite in $6$ seconds. The total distance traveled by the body during this time interval is

Open in App

Solution

Initial velocity, $\begin{matrix} u = 36 \times \frac{5}{18} = 10 m s^{1} \end{matrix}$

velocity after $\begin{matrix} 6 s = 18 \times \frac{5}{18} = 5 m s^{1} \end{matrix}$

$\begin{matrix} v = u + a t - 5 = 10 + a \times 6 a = - 2.5 m s^{1} \end{matrix}$

Let distance travelled to the right side $s_{1}$
$\begin{matrix} v^{2} = u^{2} + 2 a s_{1} 0 = 10^{2} + 2 (- 2.5) s_{1} s_{1} = 20 m \end{matrix}$

Let distance travelled to the left side be $s_{2}$
$\begin{matrix} 5^{2} = 2 \times 2.5 \times s_{2} s_{2} = 5 m \end{matrix}$

Total distance
$\begin{matrix} = 20 + 5 = 25 m \end{matrix}$

Suggest Corrections

Similar questions

10 m s^{- 2}

5

2

50 m s^{- 1}

3 s

90

36

5

3.06 m

0.34 m / s

0.18 m / s

n^{t h}

Join BYJU'S Learning Program