Question

Driver of the car applies brakes when the car is moving at a speed 20m/s and at a distance of 50m from the wall. Brakes produce a constant retardation of 5m/s². Find the distance of the car from wall at 3 sec.​

Answer 1

Answer:

• Distance of Car from the wall after 3 seconds of applying brakes will be 12.5 metres.

Explanation:

Given,

• Initial velocity of Car, u = 20 m/s
• Initial Distance of Car from the wall, d = 50 m
• Acceleration produced due to brakes, a = -5 m/s²

 [Since retardation produced is 5 m/s²]

To find,

• Final Distance of Car from the wall after time (t) of 3 sec, D = ?

Formula required,

Second equation of motion

• s = u t + 1/2 a t²

[ Where s is distance covered, u is initial velocity, t is time taken, a is acceleration ]

Solution,

Using second equation of motion

calculating distance covered (s) by Car in 3 seconds after applying brakes

→ s = u t + 1/2 a t²

→ s = ( 20 ) ( 3 ) + 1/2 ( -5 ) ( 3 )²

→ s = 60 - 45/2

→ s = 60 - 22.5

→ s = 37.5 m

so, Car covered a distance of 37.5 metres.

Now,

→ Final Distance of Car from wall, D = d - s

→ D = 50 - 37.5

→ D = 12.5 m

Therefore,

• Distance of Car from the wall after 3 seconds of applying brakes will be 12.5 metres.

Answer 2

GIVEN:

• Initial speed= u= 20m/s
• Distance from wall= d= 50m
• Acceleration = a= (-5)m/s²
• Time= t= 3sec

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Need to Find:

• The distance from wall at t= 3sec

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Solution:

Using, Newton's 2nd Law of Motion,

we get:

$s = ut + \dfrac{1}{2} a {t}^{2}$

==>$s = 20 \times3 + \dfrac{1}{2} \times( - 5) \times {3}^{2}$

==>$s = 60 - \dfrac{45}{2}$

==>$s = \dfrac{120 - 45}{2} m$

==>$s = 37.5m$

The distance travelled in 3sec is 37.5 m.

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Now the distance from the wall:

= d- s

= 50m- 37.5m

=$\bold{\red{12.5m\:(Ans)}}$