Question

A car moves with uniform
velocity of 40 ms for 5 s. It
comes to rest in the next 10 s
with uniform deceleration. Find
(1) deceleration
total
distance travelled by the car
(-4 ms 400 m
m​

Answer 1

Given:-

• Initial Velocity of the car = 40m/s

• Time moves (v¹) = 5s

• Final Velocity = 0m/s

• Time taken to Stop(v²) = 10s.

To Find:-

• Declaration of the car.

Formulae used:-

• v = u + at

• v² - u² = 2as.

Now,We will find Acceleration.

v = u + at

(0) = 40 + a × 10

We take t² because it is the time when car started retarding)

-40 = 10a

a = -40/10

a = -4m/s².

Therefore,

For the First 5 second

Distance travelled = V × t¹

Distance travelled (s¹)= 40 × 5 = 200m.

Now,

In the next 10 second.

v² - u² = 2as

(0)² - (40)² = 2 × -4 × s

-1600 = -8s

s = -1600/-8

s² = 200m

Now

Total Distance travelled = (s¹ + s²) = 200 + 200

Total Distance Travelled = 400m.

Hence, The total Distance Travelled by car is 400m