Question

a car moving with a velocity of 36 km/h . when brakes are applied it stop after 10m. calculate

acceleration and time taken to stop.​

Answer 1

Given

• Initial Velocity = 36 km/h
• Distance = 10 m
• Final Velocity = 0 m/s

To Find

• Acceleration and the time taken

Solution

● v²-u² = 2as [Third Equation of Motion]

● v = u + at [First Equation of Motion]

✭ Acceleration of the car :

→ v²-u² = 2as

• v = Final Velocity = 0 m/s
• u = Initial Velocity = 36 km/h = 36 × 5/18 = 10 m/s
• a = Acceleration = ?
• s = Distance = 10 m

→ 0²-10² = 2 × a × 10

→ -100 = 20a

→ -100/20 = a

→ Acceleration = -5 m/s²

━━━━━━━━━━━━━━━━━━━

✭ Time taken :

→ v = u + at

→ 0 = 10 + (-5) × t

→ -10 = -5t

→ -10/-5 = t

→ Time = 2 sec

Therefore,

• Acceleration of the car is -5 m/s²
• Time taken is 2 seconds

Answer 2

Answer:

Given :-

• Initial velocity (U) = 36 km/h
• Distance (S) = 10 m
• Final velocity (V) = 0 m/s

To Find :-

• Acceleration of car
• Time taken to stop

Solution :-

❶ Acceleration of car

Here we will use newton third Equation.

$\huge \mathrm { {v}^{2} = {u}^{2} + 2as}$

Here,

V = Final Velocity = 0 m/s

U = Initial velocity = 36 × 5/18 = 10 m/s

S = Distance = 10 m

A = Acceleration

$\sf \: {0}^{2} = {10}^{2} + 2(a)(10)$

$\sf \: 0 = 100 + 20(a)$

$\sf \: 0 - 100 = 20 \: a$

$\sf \: - 100 = 20a$

$\sf \: a = \dfrac{ - 100}{20}$

$\sf \: a = - 5 \: mps$

Hence,

The acceleration of car = 5 m/s.

❷ Time taken

Here, we will use newton first Equation of motion.

$\huge \mathrm{v = u + at}$

Here,

V = Final Velocity

U = Initial Velocity

A = Acceleration

T = Time

$\sf \: 0 = 10 + ( - 5)(t)$

$\sf \: 0 - 10 = - 5t$

$\sf \: - 10 = - 5t$

$\sf \: t = \dfrac{ - 10}{ - 5} = \dfrac{ - 2}{ - 1} = \dfrac{2}{1} = 2$

Hence, time taken by car is 2 second.