Question

(c) A car travels with a uniform velocity of 25 m/s for 5 s. The breaks 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 car after applying the breaks.
(
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Answer 1

Answer

• The Distance travelled by the car before the breaks are applied is 125 m
• The retardation if the body is 2.5 m/s²
• Distance travelled by the car after applying the breaks is 125 m

Explanation

Given

• A car has a uniform velocity of 25 m/s and travels for 5 sec
• Then breaks are applied and the body comes to rest in 10 seconds

To Find

• Distance travelled by the car before the breaks are applied
• The retardation if the body
• Distance travelled by the car after applying the breaks

Solution

• First use the speed equation to find the distance travelled. Then for the retardation use the first equation of motion and then the third equation for the last question

✭ Distance Covered before the breaks are applied

→ Speed = Distance/Time

→ 25 = Distance/5

→ 25 × 5 = Distance

→ Distance = 125 m

✭ Retardation of the body

→ v = u+at

• Initial Velocity = 25 m/s
• Final Velocity = 0 m/s
• Time = 10 sec

→ 0 = 25 + a×10

→ -25 = 10a

→ -25/10 = a

→ Retardation = 2.5 m/s²

[Acceleration is -2.5 m/s² but here it is Retardation and so we can here the -ve sign]

✭ Distance Travelled by the car after the breaks are applied

→ v²-u² = 2as

• u = 25 m/s
• v = 0 m/s
• a = 2.5 m/s²

→ 0² - 25² = 2 × 2.5 × s

→ 0-625 = 5×s

→ -625/5 = s

→ Distance = 125 m [Ignore Negative sign as it is -ve just because the body's speed is reducing]