Question

A motor car slows down from 72 kmh⁻¹ to 36 kmh⁻¹ over a distance of 25 m. If the brakes are applied with the same force
calculate

1. total time in which car comes to rest
2. distance travelled by it.

Answer 1

Take motion of a car from A to B:

u = 72 kmh⁻¹ = `72xx5/18` ms⁻¹
u = 20 ms⁻¹
v = 36 kmh⁻¹ = `36xx5/18` ms⁻¹
v = 10 ms⁻¹
(Time) t₁ = ?
Acceleration = a = ?
Distance = S₁ = 25 m
v² − u² = 2aS₁
(10)² − (20)² = 2a (25)
50a = 100 − 400 = −300

a = `-300/50` = −6 ms⁻²
Now, v = u + at₁
10 = 20 + (−6) t₁
6t₁ = 20 − 10 = 10
t₁ = `10/6` = 1.67s

Take motion of a car from B to C:

u = 36 kmh⁻¹ = 10 ms⁻¹
v = 0
a = −6 ms⁻²
(Time) t₂ = ?
(Distance) S₂ = ?
v = u + at₂
0 = 10 + (−6) t₂
6t₂ = 10
t₂ = `10/6` = 1.67S
Now, v² − u² = 2aS₂
(0)² − (10)² = 2 (−6) S₂
−12S₂ = −100
S₂ = `100/12` = 8.33 m

Total time taken by car from A to C = t₁ + t₂ = `10/6+10/6`

= `(10+10)/6=20/6` = 3.33s
Total distance covered by car from A to C = S₁ + S₂
= 25 + 8.33 = 33.33 m