Question

Q4. A vehicle is accelerating on a straight road. Its velocity at any instant is 30 kmhr-1, after 2s, it is 33.6 kmhr-1 and after further 2s, it is 37.2 kmhr -1. Find the acceleration in ms-2. Is the acceleration uniform?

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Acceleration=0.5\:m/s^{2}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given:}} \\ \tt: \implies Initial \: velocity(u) = 30 \: km/h \\ \\ \tt: \implies Final \: velocity( v_{1} ) = 33.6 \: km/h \\ \\ \tt: \implies Final\: velocity( v_{2} ) = 37.2 \: km/h \\ \\ \tt: \implies Time( t_{1}) = 2 \: sec \\ \\ \tt: \implies Time( t_{2}) = 2 \: sec \\ \\ \red{\underline \bold{To \: Find:}}\\ \tt: \implies Acceleration(a) = ?$

• According to given question :

$\tt \circ \: Initial \: velocity =30 \times \frac{5}{18} = \frac{150}{18} \: m/s \\ \\ \tt \circ \: Final \: velocity = 33.6 \times \frac{5}{18} = \frac{168}{18} \: m/s \\ \\ \tt \circ \: Time = 2 \: sec \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies v = u + at \\ \\ \tt: \implies \frac{168}{18} = \frac{150}{18} + a \times 2 \\ \\ \tt: \implies \frac{168 - 150}{18} = 2 \times a \\ \\ \tt: \implies a = \frac{18}{18\times 2} \\ \\ \green{\tt: \implies a = 0.5 \: m/{s}^{2} } \\ \\ \huge\bold{Again,} \\ \\ \tt \circ \:Initial \: velocity = \frac{168}{18} \: m/s \\ \\ \tt \circ \:Final \: velocity = 37.2 \times \frac{5}{18} = \frac{186}{18} \: m/s \\ \\ \tt \circ \: time = 2 \: sec \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies v = u + at \\ \\ \tt: \implies \frac{186}{18} = \frac{168}{18} + a \times 2 \\ \\ \tt: \implies \frac{186 - 168}{18} = 2 \times a \\ \\ \tt: \implies \frac{18}{18\times 2} = a \\ \\ \green{\tt: \implies a = 0.5 \: m/{s}^{2} } \\ \\ \green{\tt \therefore Acceleration \: is \: uniform \: in \: interval}$

Answer 2

$\Large{\underline{\underline{\bf{Solution :}}}}$

↪ Case : 1

$\Large{\underline{\underline{\bf{Given :}}}}$

Initial velocity of car (u) = 30 km/h = $\sf{30 \times \frac{5}{18} = \frac{150}{18} \: \: ms^{-1}}$

Final velocity($\sf{v_1})$ = 33.6 km/h = $\sf{33.6 \times \frac{5}{18} = \frac{168}{18} \: \: ms^{-1}}$

Time taken (t) = 2 seconds

$\rule{200}{1}$

$\Large{\underline{\underline{\bf{Explanation :}}}}$

As, We have to find the acceleration.

We know that,

$\Large{\implies{\boxed{\boxed{\sf{a = \frac{v_1 - u}{t}}}}}}$

★ Put Values

$\sf{\rightarrow a = \dfrac{\dfrac{168}{18} - \frac{150}{18}}{2}} \\ \\ \sf{\rightarrow a = \dfrac{ \dfrac{168 - 150}{18}}{2}} \\ \\ \sf{\rightarrow a = \dfrac{ \dfrac{\cancel{18}}{\cancel{18}}}{2}} \\ \\ \sf{\rightarrow a = \frac{1}{2}} \\ \\ \sf{\rightarrow a = 0.5 \: ms^{-2}}\\ \\ \Large{\implies{\boxed{\boxed{\sf{a = 0.5 \: ms^{-2}}}}}}$

$\rule{200}{2}$

↪ Case : 2

$\Large{\underline{\underline{\bf{Given :}}}}$

Initial velocity (u) = $\frac{168}{18}$ m/s

Final velocity ($\sf{v_2})$ = 37.2 km/h = $\sf{37.2 \times \frac{5}{18} = \frac{186}{18} \: \: ms^{-1}}$

Time taken (t) = 2 seconds

$\rule{200}{1}$

$\Large{\underline{\underline{\bf{Explanation :}}}}$

As, We have to find the acceleration.

We know that,

$\Large{\implies{\boxed{\boxed{\sf{a = \frac{v_1 - u}{t}}}}}}$

★ Put Values

$\sf{\rightarrow a = \dfrac{\dfrac{186}{18} - \frac{168}{18}}{2}} \\ \\ \sf{\rightarrow a = \dfrac{ \dfrac{186 - 168}{18}}{2}} \\ \\ \sf{\rightarrow a = \dfrac{ \dfrac{\cancel{18}}{\cancel{18}}}{2}} \\ \\ \sf{\rightarrow a = \frac{1}{2}} \\ \\ \sf{\rightarrow a = 0.5 \: ms^{-2}} \\ \\ \Large{\implies{\boxed{\boxed{\sf{a = 0.5 \: ms^{-2}}}}}}$