Question

Q-59. A jeep travels with a velocity of
108 km/h and is brought to rest by
applying the brakes. It experiences a
retardation of 3 m/s 2. The distance it
travels before coming to rest is
A
120 m
B.
130 m
С
140 m
D) 150 m​

Answer 1

u = 108km/h = 30m/s

v= 0m/s

a = 3m/s^2

s= ?

From 3rd equation of motion,

${v}^{2} = {u}^{2} + 2as$

${0}^{2} = {30}^{2} + 2(3)s$

$0 = 900+ 6s$

$- 900 \div 6 = s$

$150 = s$

Answer : D) 150m

Hope it helps!!

Answer 2

Given :-

• Initial velocity (u) = 108 km/h = 30 m/s
• Final velocity (v) = 0 m/s (It applies brakes)
• Retardation (a) = - 3 m/s²

To find :-

• Distance travelled (s)

According to the question,

By using Newtons third equation of motion,

→ v² = u² + 2as

Where,

• v stands for Final velocity
• u stand for Initial velocity
• a stands for Acceleration
• s stands for Distance

→ Substituting the values,

→ (0)² = (30)² + 2 × (-3) × s

→ 0 = 900 + (-6s)

→ 0 - 900 = - 6s

→ -900 = -6s

→ 900 ÷ 6 = s

→ 150 = s

.°. The distance travelled is 150 m.

So,option D) 150 m is correct.

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