Answer:

given : car speed 52km/ per hour

The total displacement of each car can be achieved by calculating the area beneath the speed-time graph.

Therefore, displacement of the first car = area of triangle

AOB= 1/2× (OB)×(OA)

But OB = 5 seconds and OA = 52 km.h-1-= 14.44 m/s

Therefore, the area of the triangle AOB is given by: 1/2×(5s)×(14.44ms-1) = 36 meters

Now, the displacement of the second car is given by the area of the triangle

COD= 1/2×(OD)×(OC)

But OC = 10 seconds and OC = 3km.h-1= 0.83 m/s

Therefore, area of triangle COD = 12× (10s)×(0.83ms-1) = 4.15 meters

Hence, the first car is displaced by 36 meters whereas the second car is displaced by 4.15 meters.

Therefore, the first car (which was travelling at 52 kmph) travelled farther post the application of brakes

hope it helps you ✅✅✅

Required Answer :-

According to the question,

★ Given -

Case I ( Car A )

• Initial Velocity for driver A - 52km h-1
• Time - 5 s

Case 2 ( Car B )

• Initial Velocity for driver A - 3km h-1
• Time - 10 s

So, firstly we will convert the initial velocity of both car in m/s

★ In order to convert it into metre per second we have to multiply it with 5/18

• For car A -

• For car B -

On the same graph paper, plot the speed Vs time graphs for the two cars.

We know that

• For car A

• Initial velocity = 14.44 m/s
• Time = 5s
• Final velocity = 0

• For car B

• initial velocity = 0.83 m/s
• Time = 10s
• Final velocity = 0

So, let's plot the graph

(see the attachment)

Which of the the both cars travelled further after the breakers were applied ?

★ For Car A -

Given -

• u = 14.44 m/s
• t = 5 s
• v = 0

Distance - Area under VT graph

★ For Car B -

Given -

• u = 0.83
• t = 10 s
• v = 0

Distance - Area under VT graph

☀️ Hence Car A travelled further after the breakers were applied !