Question

On application of broke to a car a negative acceleration of 5m/s is produced. If car takes 3 sec to stop after applying brakes . Find distance traveled by it during this time.​

Answer 1

Answer :

Given that on application of broke to a car of negative acceleration of 5 m/s² is produced and the car takes 3 s to stop.

We have to find distance travelled by it during this time.

From the data :

- Negative acceleration = 5 m/s²

- Acceleration,a = -5 m/s²

- Time, t = 3 s

- Final speed, v = 0 m/s [As it stops]

We will use 1st equation of motion to find initial speed of car :

→ v = u + at

→ 0 = u + (-5)3

→ 0 = u - 15

→ u = 15 m/s

Now we will use 3rd equation of motion to find distance travelled by car :

→ v² = u² + 2as

→ 0² = 15² + 2(-5)s

→ 0 = 225 - 10s

→ 10s = 225

→ s = 225/10

→ s = 22.5 m

∴ Distance travelled by car = 22.5 m ✓

Answer 2

Answer:

$\boxed{\sf Distance \: travelled \: by \: car \: after \: applying \: brakes = 22.5 \: m}$

Given:

Negative acceleration (Retardation) of car = 5 m/s² OR

Acceleration = -5 m/s²

Time taken by car to stop after applying brakes = 3 sec

Final speed of car = 0 m/s (As it stops after applying brakes)

To find:

Distance travelled by car after applying brakes

Explanation:

Let the distance travelled by car after applying brakes = s

First we need to find the initial speed;

For that we will use $\sf 1^{st}$ equation of motion:

$\sf \implies v = u + at \\ \\ \sf \implies 0 = u + ( - 5)(3) \\ \\ \sf \implies 0 = u + ( - 15) \\ \\ \sf \implies 0 = u - 15 \\ \\ \sf \implies u - 15 = 0 \\ \\ \sf \implies u = 15 \: m/s$

Now we can calculate distance travelled by car after applying the brakes using $\sf 3^{rd}$ equation of motion:

$\sf \implies {v}^{2} = {u}^{2} + 2as \\ \\ \sf \implies {(0)}^{2} = {(15)}^{2} + 2( - 5)s \\ \\ \sf \implies 0 = 225 + ( - 10)s \\ \\ \sf \implies 0 = 225 - 10s \\ \\ \sf \implies 225 - 10s = 0 \\ \\ \sf \implies \cancel{- }10s = \cancel{- }225 \\ \\ \sf \implies 10s = 225 \\ \\ \sf \implies s = \frac{225}{10} \\ \\ \sf \implies s = 22.5 \: m$

So,

Distance travelled by car after applying brakes = 22.5 m