Question

A boy is moving with a velocity of 12m/s and it co8 to rest in 18m . What was the accularation?

Answer 1

Correct Question :-

A boy is moving with initial velocity of 12m/s and it comes to rest in 18m. What is the acceleration of boy ?

Solution :-

• Initial velocity = 12mps
• Distance covered = 18m
• Final velocity = zero

→ Acceleration is defined as ratio of change in velocity to the time interval.

→ As per second equation of kinematics, relation between initial velocity, final velocity, acceleration and distance is given by

$\underline{\boxed{\bf{\pink{v^2-u^2=2as}}}}$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• s denotes distance

$\implies\sf\:v^2-u^2=2as\\ \\ \implies\sf\:(0)^2-(12)^2=2(a)(18)\\ \\ \implies\sf\:-144=36a\\ \\ \implies\sf\:a=-\dfrac{144}{36}\\ \\ \implies\boxed{\bf{\orange{a=-4\:ms^{-2}}}}$

→ Negative sign shows retardation.

Answer 2

Given ,

• Initial velocity (u) = 12 m/s
• Final velocity (v) = 0
• Distance (s) = 18 m

We know that ,

The Newton's third equation of motion is given by

$\mathtt{ \large \fbox{ {(v)}^{2} - {(u)}^{2} = 2as}}$

Substitute the known values , we get

➡(0)² - (12)² = 2 × a × 18

➡- 144 = 36 × a

➡a = -144/36

➡a = -4 m/s²

The negative sign of acceleration indicates it is redaration

Hence , the redaration is -4 m/s²