A car cruises at a constant speed of 300 ms through a distan

A car cruises at a constant speed of 30.0 m/s through a distance of 300 m when the brakes are so that the car decelerates at a rate of 0.68 m/s^2 for 3.5 s. What is the total distance traveled by the car?

Solution

constant speed, V1 = 30m/s.

Distance ,D = 300 m.

=> Total time will be , T1= D/V1 = 300/30 = 10s.

But in the question it is given that when the brakes are applied , the car decelerates at a rate of 0.68 m/s² for 3.5s.

Therefore,   Deceleration is, A_d = 0.68 m/s^2.

                 For time, T_d =3.5 s.

    as we know, A_{d =} S_d/ T_d

               => 0.68 = S_d/3.5

               => S_d = 2.38 m/s.

Constant speed decreases by the amount of 2.38 m/s due to decelaration.

Therefore, the speed for 3.5 s due to decelaration would be V2 = V1-2.38 = 30-2.38 = 27.62m/s.

Distance travelled by the car at a speed of 27.62m/s for 3.5s is D1= 96.97 m .

Time remaining is 10-3.5 = 6.5 s (It is the time during which the car is travelling with constant speed of 30 m/s)

Now, distance travelled by the car at a speed of 30 m/s for 6.5s is D2= 195 m.

Therefore, total distance travelled by the car due to deceleration would be

D =D1+D2 = 96.67+195 = 291.67 m (is the required answer).