Question

A car starts from rest with a uniform acceleration of 0.5 m/s2. What will its velocity be after it has covered 25 m? How long did it take to cover this distance? *

Answer 1

GiveN :

• Initial velocity (u) = 0 m/s
• Acceleration (a) = 0.5 m/s²
• Distance (s) = 25 m

To FinD :

• Final velocity when it travels 25m .
• Time taken to travel 25 m

SolutioN :

$\underbrace{\sf{Velocity \: when \: it \: travels \: 25 \: m}}$

$\implies \sf{v^2 \: - \: u^2 \: = \: 2as} \\ \\ \implies \sf{v^2 \: - \: 0^2 \: = \: 2(0.5)(25)} \\ \\ \implies \sf{v^2 \: = \: 25} \\ \\ \implies {\sf{v \: = \: \sqrt{25}}} \\ \\ \implies \sf{v \: = \: 5} \\ \\ \underline{\sf{\therefore \: Final \: velocity \: is \: 5 \: ms^{-1}}}$

_____________________________

$\underbrace{\sf{Time \: taken \: to \: travel \: 25 \: m}}$

$\implies \sf{v \: = \: u \: + \: at} \\ \\ \implies \sf{5 \: = \: 0 \: + \: 0.5t} \\ \\ \implies \sf{t \: = \: \dfrac{5}{0.5}} \\ \\ \implies \sf{t \: = \: \dfrac{50}{5}} \\ \\ \implies \sf{t \: = \: 10} \\ \\ \underline{\sf{\therefore \: Time \: taken \: to \: travel \: 25 \: m \: is \: 10s}}$

Answer 2

Given:

Initial velocity = 0 m/s

Acceleration = 0.5 m/s²

Distance travelled = 25 m

To find:

Final velocity of the object

Concept:

Since the acceleration is constant , we can easily apply equation of kinematics to solve this question

Calculation:

Let final velocity be v , initial velocity be u, acceleration be a , distance be s.

Applying Equation of Kinematics :

${v}^{2} = {u}^{2} + 2as$

$= > {v}^{2} = {(0)}^{2} +( 2 \times 0.5 \times 25)$

$= > {v}^{2} = 25$

$= > v = \sqrt{25}$

$= > v = 5 \: m {s}^{ - 1}$

Final velocity is 5 m/s

Let time taken be t :

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

$=> t = \dfrac{5-0}{0.5}$

$=> t = 10 sec$

So time taken is 10 sec