Question

You're arguing over a cell phone while trailing an unmarked police car by 25 m; both your car and the police car are travelling at 110 Km/h. Your argument diverts your attention from the police car for 2.0 s. At the beginning of that 2.0s, the police officer starts braking suddenly at 5 ms-².
(a) What is the separation between the two cars when your attention finally returns.
Suppose that you take another 0.40 s to realise your danger and begin braking.
(b) If you too take 5 ms-², What is your speed when you hit the police car.



Please Solve this question....​

Answer 1

Given:-

• Initial Velocity of my Car = 110km/h → 30.55m/s

• Initial Velocity of Police car = 30.55m/s

• Time given = 2s

• Retardation of Police Car = -5m/s²

To Find:-

• What is the separation between the two cars when your attention finally returns. Suppose that you take another 0.40 s to realise your danger and begin braking.

• If you too take 5 ms-², What is your speed when you hit the police car.

Formulae used:-

• Distance Covered = u × t

• v = u + at

• v² - u² = 2as

• s = ut + ½ × a × t²

Now,

According to First Question

→ t = 2 + 0.4 = 2.4s

So,

→ S covered by my Car = u × t

→ 30.55 × 2.4

→ 73.32m

Hence, The Distance Covered by Car us 73.32 in that 2s.

Now,

For Police car

→ v = u + at

→ v = 30.55 + (-5) × 2.4

→ v = 30.55 - 12

→ v = 18.5m/s

Therefore,

→ v² - u² = 2as

→ (18.5)² - (30.5)² = 2 × -5 × s

→ 342.25 - 930.25 = -10s

→ -588 = -10s

→ s = -588/10

→ s = 58.88m

Hence, The Distance Covered by the Police car is 58.88m

Therefore,

→ Seperation between two cars = ( 25 - 73.32 + 58.88)

→ Seperation = 10.6m

Therefore, They were seperated to each other by 10.6m

2.)

We know Collision occur at t when x¹ = x²

We also know that,

The Distance between my Car and Police car is 10.6m

So,

It should be noted that we will choose t¹ = 0 to be the reference Position this means that Position of x¹ = 10.6 ( Police )

→ x¹ - 10.6 = u( t - t¹) + ½ × a × ( t² - t¹ )

→ x¹ - 10.6 = 18.5t + ½ × -5 × t²

→ x¹ - 10.6 = 18.5t -2.5t²

→ x¹ = 10.6 + 18.5t - 2.5t²

And,

→ x = u ( t - t¹)+ ½ × a × ( t² - t¹ )

→ x = 30.56 × t + ½ × -5 × t²

→ x = 30.56t - 2.5t²

Now,

→ x¹ = x²

→ 10.6 + 18.5t - 2.5t² = 30.56t - 2.5t²

→ - 30.56t + 18.5t = -10.6

→ 30.56t - 18.5t = 10.6

→ (30.56 - 18.5)t = 10.6

→ 12.06t = 10.6

→ t = 12.06/10.6

→ t = 0.883s

Therefore,

→ v = u + at

→ v = 30.56 + -5 × 0883

→ v = 30.56 - 4.415

→ v = 26.135m/s Approx

Hence, The speed of ny car will be 26.135m/s