Question

A body is started from rest with acceleration 4 m/ sCalculate the distance travelled by body is 5th second- (1) 50 m (2) 18 m (3) 32 m (4) 60 m​

Answer 1

Answer:

Here is your answer

Explanation:

This may help you

Answer 2

$\small{{\orange{ \boxed{ \boxed{ \begin{array}{cc} \bf \to \: given \: : \\ \\ \sf \hookrightarrow \: initial \: velocity \: \: u = 0 \: \: \{ \: \because \: at \: rest \} \\ \\ \sf \hookrightarrow \: acceleration \: \: a = 4 \: m {s}^{ - 2} \\ \\ \sf \hookrightarrow \: time \: \: t = 5\: s \: \\ \\ \blue{ \underline{ \text \: we \: have \: to \: find \: : }} \\ \\ \sf \hookrightarrow \: distanse \: travelled \: by \: {5}^{th} sec \: \: is \: = \: S \\ \\ \\ \red{ \underline{ \bf \: solution }} \\ \\ \sf \: we \: know \: that \\ \\ \blue{ \boxed{\sf \: distance \: travelled \: by \: {t}^{th} \: s \: is \:: S= u + \frac{1}{2}a(2t - 1) }} \\ \\ \\ \therefore \: \sf\:distance \: travelled \: by \: {5}^{th}sec \: is \: : \\ \\ \sf \: S = 0 + \frac{1}{2} \times 4(2 \times 5 - 1) \\ \\ = \frac{1}{2} \times 4(10 - 1) \\ \\ = 2 \times 9 \\ \\ = 18 \: m\end{array}}}}}}$

So option (2) 18 m is the correct answer