Question

Determine 'a' of the object which
(a) moves in a straight line with a constant speed of
20 m s- for 12 seconds.
(b) changes its velocity from 0 m h-1 to 360 m min-1
in 4.2 s.
​

Answer 1

$\huge \underline {\underline{ \mathfrak{ \green{Ans}wer \colon}}}$

(a) Object is moving in a straight line with constant speed of 20 m/s in 12 s.

So,

$\sf Given \begin{cases} \rm{Initial \:Velocity \: (u) \: = \: 20 \: ms^{-1}} \\ \rm{Final \: Velocity \: (v) \: = \: 20 \: ms^{-1}} \\ \rm{Time \: (t) \: = \: 12 \: s} \\ \rm{Acceleration \: (a) \: = \: ?} \end{cases}$

Use formula for Acceleration :

$\large{\boxed{\sf{a \: = \: \dfrac{v \: - \: u}{t}}}} \\ \\ \implies {\sf{a \: = \: \dfrac{20 \: - \: 20}{12}}} \\ \\ \implies {\sf{a \: = \: \dfrac{0}{12}}} \\ \\ \implies {\sf{a \: = \: 0}} \\ \\ {\underline{\sf{\therefore \: Acceleration \: is \: 0 \: ms^{-2}}}}$

$\rule{200}{2}$

(b) Object changes its velocity from 0 m/h to 360 m/s

$\sf Given \begin{cases} \rm{Initial \:Velocity \: (u) \: = \: 0 \: ms^{-1}} \\ \rm{Final \: Velocity \: (v) \: = \: 360 \: mm^{-1} \: = \: 6 \: ms^{-1}} \\ \rm{Time \: (t) \: = \: 4.2 \: s} \\ \rm{Acceleration \: (a) \: = \: ?} \end{cases}$

Use same formula :

$\large{\boxed{\sf{a \: = \: \dfrac{v \: - \: u}{t}}}} \\ \\ \implies {\sf{a \: = \: \dfrac{6 \: - \: 0}{4.2}}} \\ \\ \implies {\sf{a \: = \: \dfrac{6}{4.2}}} \\ \\ \implies {\sf{a \: = \: 1.42}} \\ \\ \underline{\sf{\therefore \: Acceleration \: is \: 1.42 \: ms^{-2}}}$

Answer 2

Answer:

Given:

1. Object moves at constant speed at 20 m/s for 12 seconds.
2. Changes it's Velocity from 0 m/hr to 360 m/min

To find:

Acceleration in both cases

Calculation:

$\green{ \sf{ \huge{ \underline{1st \: Case : }}}}$

Here the car moves with constant Velocity. This means that the Velocity hasn't changed with respect with time.

Hence acceleration is zero.

Acceleration = 0 m/s².

$\green{ \sf{ \huge{ \underline{2nd \: Case : }}}}$

First you need to convert the units of Velocity to a single type.

Initial Velocity = 0 m/hr = 0 m/s

Final Velocity = 360 m/min

=> Final Velocity = 360 × (1/60)

=> Final Velocity = 6 m/s

So, acceleration be a :

$a = \dfrac{(6 - 0)}{4.2} =1.428 \: m {s}^{ - 2}$