Question

 A car starts from rest and accelerates uniformly to a speed of 72 km/h. over a distance of 500 m. A further acceleration raises the speed to 90 km/h. in 10 seconds. The brakes are now applied to bring the car to rest under uniform retardation in 5 seconds. Find the distance travelled during braking.​

Answer 1

Answer:

Given

u=0

v=72 km/h=20 m/s

t=2s

a=

t

v−u

a=2

s=ut+

2

at

2

putting values

s=100m

Answer 2

Answer:

225 metres

Step-by-Step Explanation :

Given:

initial velocity, u = 0

final velocity, v = 72 km/h = 20 m/s

distance, s = 500 m

Firstly,

let us find the motion of the car from rest.

Let, a = Acceleration of the car

As we know that, according to third equation of motion,

$\rm v^2 = u^2 + 2as$

Putting the values,

$\rm \therefore (20)^2 = (0)^2 + 2a \times 500 \\ \\ \implies \rm 400 = 2a \times 500 \\ \\ \implies \rm 2a = \frac{400}{500} \\ \\ \implies \rm 2a = \frac{4}{5} \\ \\ \implies \rm a = \frac{4}{5} \times \frac{1}{2} \\ \\ \implies \rm a = \frac{4}{10} \\ \\ \implies \rm \bf \red{a = 0.4m/s^2}$

So,

Lets take t = Time taken by car to attain the speed.

As, we know that, according to first equation of motion,

$\rm v = u + a.t$

Putting the values,

$\therefore \rm 20 = 0 + 0.4 \times t \\ \\ \implies \rm t = \frac{20}{0.4} \\ \\ \implies \rm \bf \red{t = 50 s}$

Now,

Let's find the motion of the car from 72 km/h to 90 km/h in 10 seconds.

Given :

Initial velocity, u = 72 km/h = 20 m/s

Final velocity, v = 96 km/h = 25 m/s

Time, t = 10 s

Take a = Acceleration of the car.

As, We know that, according to the first equation of motion,

$\rm v = u + a.t$

Putting the values,

$\rm \therefore 25 = 20 + a \times 10 \\ \\ \implies \rm 25 - 20 = a \times 10 \\ \\ \implies \rm 5 = a \times 10 \\ \\ \implies \rm a = \frac{5}{10} \\ \\ \implies \rm \bf \red{a = 0.5 m/s^2}$

To find the distance moved by the car :

As we know that, according to the third equation of motion,

$s = ut + \frac{1}{2} at^2$

Putting the values,

$\therefore \rm s = 20 \times 10 + \frac{1}{2} \times 0.5 \times (10)^2 \\ \\ \implies \rm 200 + \frac{0.5 \times 100}{2} \\ \\ \implies \rm 200 + \frac{50}{2} \\ \\ \implies \rm 200 + 25 \\ \implies \rm \bf \blue{s = 225m}$

Hence, the distance travelled by the Car during braking is 225 meters.