Question

A body is moving with velocity of 20 m/s, after 5 s its velocity changes to 40 m/s. The acceleration of the body will be​

Answer 1

Answer:

4

Explanation:

Given, initial velocity , u = 20 m/s

Final Velocity, v = 40 m/s

time, t = 5 s

Acceleration is given by,

$a =\frac{v-u}{t}$

$a = \frac{40 - 20}{5}$

$a = \frac{20}{5}$

$a = 4 m/s^{2}$

Answer 2

Provided that:

• Initial velocity = 20 mps
• Time = 5 seconds
• Final velocity = 40 mps

To calculate:

• Acceleration

Solution:

• The acceleration = 4 m/s²

Using concept(s):

• To solve this question we can use either first equation of motion or acceleration formula!

• Choice may vary!

Using formula(s):

• Acceleration formula:

• ${\small{\underline{\boxed{\pmb{\sf{a \: = \dfrac{v-u}{t}}}}}}}$

• First equation of motion:

• ${\small{\underline{\boxed{\pmb{\sf{v \: = u \: + at}}}}}}$

Where, v denotes final velocity, u denotes initial velocity, a denotes acceleration and t denotes time taken.

Required solution:

~ Let us find out the acceleration by using acceleration formula first!

$:\implies \sf a \: = \dfrac{dv}{dt} \\ \\ :\implies \sf a \: = \dfrac{\Delta \: v}{t} \\ \\ :\implies \sf a \: = \dfrac{Change \: in \: velocity}{Time} \\ \\ :\implies \sf a \: = \dfrac{v - u}{t} \\ \\ :\implies \sf a \: = \dfrac{40-20}{5} \\ \\ :\implies \sf a \: = \dfrac{20}{5} \\ \\ :\implies \sf a \: = \cancel{\dfrac{20}{5}} \: (Cancelling) \\ \\ :\implies \sf a \: = 4 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = 4 \: ms^{-2}$

~ Now let us calculate acceleration by using first equation of motion!

$:\implies \sf v \: = u \: + at \\ \\ :\implies \sf 40 = 20 + a(5) \\ \\ :\implies \sf 40 - 20 = 5a \\ \\ :\implies \sf 20 = 5a \\ \\ :\implies \sf a \: = \dfrac{20}{5} \\ \\ :\implies \sf a \: = \cancel{\dfrac{20}{5}} \: (Cancelling) \\ \\ :\implies \sf a \: = 4 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = 4 \: ms^{-2}$

• Choice may vary!

Explore more:

About acceleration:

$\begin{gathered}\boxed{\begin{array}{c}\\ {\pmb{\sf{What \: is \: acceleration?}}} \\ \\ \sf The \: rate \: of \: change \: of \: velocity \: of \: an \\ \sf object \: with \: respect \: to \: time \\ \sf is \: known \: as \: acceleration. \\ \\ \sf \star \: Negative \: acceleration \: is \: known \: as \: Deceleration. \\ \sf \star \: Deceleration \: is \: known \: as \: retardation. \\ \sf \star \: It's \: SI \: unit \: is \: ms^{-2} \: or \: m/s^2 \\ \sf \star \: It \: may \: be \: \pm ve \: or \: 0 \: too \\ \sf \star \: It \: is \: a \: vector \: quantity \\ \\ {\pmb{\sf{Conditions \: of \pm ve \: or \: 0 \: acceleration}}} \\ \\ \sf [1] \: Positive \: acceleration: \: \sf When \: \bf{u} \: \sf is \: lower \: than \: \bf{v} \\ \sf [2] \: Negative \: acceleration: \: \sf When \: \bf{v} \: \sf is \: lower \: than \: \bf{u} \\ \sf [3] \: Zero \: acceleration: \: \sf When \: \bf{v} \: \sf and \: \bf{u} \: \sf are \: equal \end{array}}\end{gathered}$

Equations of motion:

$\begin{gathered}\boxed{\begin{array}{c}\\ {\pmb{\sf{Three \: equations \: of \: motion}}} \\ \\ \sf \star \: v \: = u \: + at \\ \\ \sf \star \: s \: = ut + \: \dfrac{1}{2} \: at^2 \\ \\ \sf \star \: v^2 - u^2 \: = 2as\end{array}}\end{gathered}$