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

Given that, retardation of 2m/s is produced when brakes are applied to a running car. After 5 sec it stops.

Here, Retardation (a) = -2 m/s² and time (t) = 5 sce

Let us assume that car us running with a speed of u m/s or with an initial velocity of u m/s.

After applying brakes, final velocity i.e. v become 0 m/s.

Using the First Equation of Motion,

v = u + at

0 = u + (-2)(5)

-u = -10

u = 10 m/s

Therefore, the initial velocity of the car is 10 m/s.

We have to find the distance (s) travelled by car after brakes are applied.

Using the Second Equation of Motion,

s= ut + 1/2 at²

s= 10(5) + 1/2 × (-2)(5)²

s= 50 + (-1)(25)

s= 50 - 25

s= 25

Therefore, the distance covered by the car is 25 m.

Answer 2

Answer:

• Retardation (a) = – 2 m/s²
• Time (t) = 5 s
• Final Velocity (v) = 0 m/s (As Car Stops at the end)

☯ By the First Equation of Motion :

⇴ v = u + at

• Putting values of each

⇴ 0 = u + ( – 2) m/s² × 5 s

⇴ 0 = u – 10 m/s

⇴ u = 10 m/s

━━━━━━━━━━━━━

☯ By the Third Equation of Motion :

⇢ v² – u² = 2as

⇢ ( 0)² – ( 10)² = 2 × ( – 2) × s

⇢ 0 – 100 = – 4s

⇢ – 100 = – 4s

• Dividing both term by – 4

⇢ s = 25 m

∴ Distance covered after applying the brakes by the car is 25 metre.