Question

A car starting from rest is accelerated for 1m/s^2 for 4 seconds. It continues at a constant velocity for 10 seconds more and then decelerates in 3 seconds until it stops. Find the total distance traveled by the car.

Answer 1

Given data : A car starting from rest is accelerated for 1 m/s² for 4 seconds. It continues at a constant velocity for 10 seconds more and then decelerates in 3 seconds until it stops.

To find : The total distance traveled by the car.

Solution : Firstly we have to calculate distance travele by car in 4 sec.

Let, initial velocity (u) of the car be 0.

We use formula of kinematical equation.

⟹ s = ut + ½ at²

Where,

• s = diplacement
• u = initial velocity {v = final velocity}
• t = time taken
• a = acceleration

Displacement for 4 sec ?

⟹ s = ut + ½ at²

⟹ s = 0 * 4 + ½ * 1 * 4²

⟹ s = 0 + ½ * 16

⟹ s = 8 m ----{1}

Velocity of car for 4 sec ?

⟹ v = u + at

⟹ v = 0 + 1 * 4

⟹ v = 4 m/s

Now, to find distance cover by car in 10 sec at constant speed.

⟹ velocity = displacement/time

⟹ 4 = displacement/10

⟹ displacement = 4 * 10

⟹ displacement = 40 m. ----{2}

Car decelerates in 3 seconds until it stops.

To find deceleration ;

⟹ v = u + at

⟹ 4 = 0 + a * 3

⟹ 4 = 3a

⟹ a = 4/3

⟹ a = 1.33 m/s²

Displacement for 3 sec ?

⟹ s = vt - ½ at²

⟹ s = 4 * 3 - ½ * 1 33 * 3²

⟹ s = 12 - ½ * 1 .33 * 9

⟹ s = 12 - ½ * 11.97

⟹ s = 12 - 5.985

⟹ s = 6.015 m ----{3}

Now total distance traveled by the car from {1} {2} and {3}

⟹ Total distance traveled by the car

= { 8 + 40 + 6.015 } m

⟹ Total distance traveled by the car

= { 8 + 40 + 6.015 } m

⟹ Total distance traveled by the car

= 54.015 m

Answer : Total distance traveled by the car is 54.015 m