A Moto Car Accelerate For 10sec to attain a velocity for 20m/s. it continue with uniform velocity for a further 20sec and the decelerate so that it stop in 2osec, calculate (1) acceleration (2) deceleration (3) the distance
To calculate the acceleration, deceleration, and distance, we can use the kinematic equations.
(1) Acceleration:
Acceleration (a) is the change in velocity (v) divided by the change in time (t).
Given:
Initial velocity (u) = 0 m/s (since the car starts from rest)
Final velocity (v) = 20 m/s
Time (t) = 10 sec
Using the formula:
a = (v - u) / t
a = (20 - 0) / 10
a = 2 m/s^2
Therefore, the acceleration of the car is 2 m/s^2.
(2) Deceleration:
Deceleration is the negative acceleration. Since the car decelerates to a halt, its final velocity will be 0 m/s.
Given:
Initial velocity (u) = 20 m/s
Final velocity (v) = 0 m/s
Time (t) = 20 sec
Using the formula:
a = (v - u) / t
a = (0 - 20) / 20
a = -1 m/s^2
Therefore, the deceleration of the car is -1 m/s^2.
(3) Distance:
To calculate the distance, we need to use the formula that relates initial velocity, final velocity, acceleration, and distance:
d = (v^2 - u^2) / (2a)
When the car is accelerating:
Given:
Initial velocity (u) = 0 m/s
Final velocity (v) = 20 m/s
Acceleration (a) = 2 m/s^2
Using the formula:
d1 = (20^2 - 0^2) / (2 * 2)
d1 = 200 / 4
d1 = 50 meters
When the car is decelerating:
Given:
Initial velocity (u) = 20 m/s
Final velocity (v) = 0 m/s
Acceleration (a) = -1 m/s^2
Using the formula:
d2 = (0^2 - 20^2) / (2 * -1)
d2 = -400 / -2
d2 = 200 meters
Therefore, the total distance traveled by the car is 50 meters + 200 meters = 250 meters.