A car starting from rest is accelerated for 1.5 m/s² for 7 seconds. It continues at a constant velocity for 20 seconds more and then decelerates in 5 seconds until it stops. Find the total distance traveled by the car.
Answers
Given data : A car starting from rest is accelerated for 1.5 m/s² for 7 seconds. It continues at a constant velocity for 20 sec more and then decelerates in 5 seconds until it stops.
To find : Find the total distance traveled by the car.
Solution : Here, according to the question, initial velocity (u) of the car is 0.
Firstly we have to calculate distance travele by car in 7 sec.
Now, by kinematical equation.
➜ s = ut + 1/2 * at²
Where,
• s = displacement
• u = initial velocity {v = final velocity}
• t = time taken
• a = acceleration
Now, with acceleration 1.5 m/s², the displacement of the car in 7 sec :
➜ s = ut + 1/2 * at²
➜ s = 0 * 7 + 1/2 * 1.5 * 7²
➜ s = 1/2 * 1.5 * 49
➜ s = 1/2 * 73.5
➜ s = 36.75 m ----{1}
Now, with acceleration 1.5 m/s², velocity of the car in 7 sec :
➜ v = u + at
➜ v = 0 + 1.5 * 7
➜ v = 10.5 m/s
A/C to the question the car moves with constant velocity (10.5 m/s) for 20 sec,
➜ Velocity = Displacement/Time
➜ 10.5 = Displacement/20
➜ Displacement = 10.5 * 20
➜ Displacement = 210 m ----{2}
Car decelerates in 5 seconds until it stops.
To find deceleration ;
➜ v = u + at
➜ 10.5 = 0 + a * 5
➜ 10.5 = 5a
➜ a = 10.5/5
➜ a = 2.1 m/s²
Now, with acceleration 2.1 m/s², the displacement of the car in 5 sec :
➜ s = vt - 1/2 * at²
➜ s = 10.5 * 5 - 1/2 * 2.1 * 5²
➜ s = 52.5 - 1/2 * 2.1 * 25
➜ s = 52.5 - 1/2 * 52.5
➜ s = 52.5 - 26.25
➜ s = 26.25 m ----{3}
Now total distance traveled by the car : [from {1} {2} and {3}]
➜ Total distance traveled by the car
= { 36.75 + 210 + 26.25 } m
➜ Total distance traveled by the car
= { 246.75 + 26.25 } m
➜ Total distance traveled by the car
= 273 m
Answer : Hence, total distance traveled by the car is 273 m.