Question

A car travels with a uniform velocity of 25 ms for
5 s. The brakes are then applied and the car is
uniformly retarded and comes to rest in further
10 s. Find : (i) the distance which the car travels
before the brakes are applied, (ii) the retardation,and (iii) the distance travelled by the car after applying the brakes​

Answer 1

GIVEn

A car travels with a uniform velocity of 25 ms for 5 s. The brakes are then applied and the car is uniformly retarded and comes to rest in further 10 s.

TO FINd

Find : (i) the distance which the car travelsbefore the brakes are applied

(ii) the retardation,and

(iii) the distance travelled by the car after applying the brakes

SOLUTIOn

(i) the distance which the car travels before the brakes are applied,

• Speed = 25m/s
• time = 5s
• distance = ?

Apply formula

→ Distance = speed × time

→ Distance = 25 × 5

→ Distance = 125m

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(ii) the retardation

• Initial velocity (u) = 25m/s
• Final velocity (v) = 0
• Time taken (t) = 10s
• Acceleration (a) = ?

→ v = u + at

→ 0 = 25 + 10a

→ 10a = -25

→ a = -2.5m/s²

_____________________

(iii) the distance travelled by the car after applying the brakes

• Final velocity (v) = 0
• Initial velocity (u) = 25m/s
• Acceleration (a) = - 2 .5 m/s²
• Distance (s) = ?

→ v² - u² = 2as

→ (0)² - (25)² = 2 × (-2.5) × s

→ 0 - 625 = -5s

→ -5s = -625

→ s = 625/5 = 125 m

_____________________

Answer 2

Given :

• A car travels at a uniforms velocity of 25 m/s for 5 sec
• Then Breaks are applied and it comes to rest in 10 s

To Find :

• Distance travelled before breaks are applied
• Value of retardation
• Distance travelled by car after applying breaks

Solution :

✯ 1) Distance Travelled before breaks are applied

• Velocity = 25 m/s
• Time = 5 s

➠ Velocity = Distance/Time

⇒ Distance = Velocity * Time

⇒Distance = 25*5

⇒Distance = 125

$\therefore$ Distance Travelled before applying breaks is 125 m

_______________________________

✯ Value of Retardation

• Initial velocity (u) = 25 m/s
• Final velocity (v) = 0 m/s
• Time (t) = 10 s

Use 1st equation of Kinematics

➠ v = u + at

⇒0 = 25 + 10a

⇒a = -25/10

⇒a = -2.5

$\therefore$ Retardation is -2.5 m/s²

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3) Distance Travelled after applying breaks

• Initial velocity = 25 m/s
• Final velocity = 0 m/s
• Acceleration = -2.5 m/s²

Use 3rd equation of Kinematics

➠ v² - u² = 2as

⇒0² - 25² = 2(-2.5)s

⇒-625 = -5s

⇒s = 625/5

⇒s = 125

$\therefore$ Distance Travelled by the car after applying breaks is 125 m