Question

A retardation of 2m/s is produced when brakes are applied to a running car.
After 5s it stops. Find the distance travelled by it after brakes are applied

Answer 1

Answer:

25m

Explanation:

we know,

v=u-at

or, 0=u-2*5

or, u=10.......(i)

now,

s=ut-1/2 at²

=10*5-1/2*2*25 [u=10 from eq.(i) ]

=50-25

=25

so,

the distance travelled by it after brakes are applied is 25 m.

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Answer 2

Given that retardation of 2 m/s² is produced when brakes are applied to a running car & after 5 seconds it stops.

We have to find distance travelled by car after brakes are applied.

Solution

As the car stops after 5 seconds, so final velocity i.e. v = 0 m/s

Again, retardation is 2 m/s²

[As, retardation shows negative ]

∴ Acceleration (a) = -2 m/s²

Using first equation of motion :

➪ v = u + at

➪ 0 = u + (— 2) × 5

➪ 0 = u - 10

➪ u = 10 m/s

∴ Initial velocity of car = 10 m/s

Using 3rd equation of motion :

➪ v² — u² = 2as

➪ 0² — 10² = 2(-2) × s

➪ 0 - 100 = -4s

➪ - 100 = -4s

➪ s = -100/-4

➪ s = 25 m

∴ Distance travelled by car after brakes applied = 25 m