A motor car accelerates for 10 seconds to attain a velocity of
20
m
/
s
2
. It continues with uniform velocity for a further 20 seconds and then decelerates so that it steps in 20 seconds. Calculate the acceleration and deceleration.
twice as long to stop, so 1/2 the acceleration
v = at, so
20/10 m/s^2, then 20 m/s, then -20/20 m/s^2
To calculate the acceleration and deceleration, we can use the equations of motion. Let's break down the motion of the motor car into three parts:
1. Acceleration: The motor car accelerates for 10 seconds to reach a velocity of 20 m/s.
For this part, we can use the equation of motion:
v = u + at
Where:
v = final velocity (20 m/s)
u = initial velocity (0 m/s)
a = acceleration
t = time taken (10 seconds)
Rearranging the equation, we get:
a = (v - u) / t
Substituting the given values, we have:
a = (20 m/s - 0 m/s) / 10 s
a = 20 m/s / 10 s
a = 2 m/s²
So, the acceleration of the motor car during the first 10 seconds is 2 m/s².
2. Uniform Velocity: The motor car continues with a uniform velocity for 20 seconds.
Since the velocity remains constant, the acceleration is zero (a = 0 m/s²).
3. Deceleration: The motor car decelerates over 20 seconds and comes to a rest.
Similar to the first part, we can use the equation of motion:
v = u + at
Where:
v = final velocity (0 m/s)
u = initial velocity (constant velocity attained during uniform motion)
a = deceleration
t = time taken (20 seconds)
Rearranging the equation, we get:
a = (v - u) / t
Since the final velocity is 0 m/s, the equation becomes:
0 = u - a * t
Substituting the given values, we have:
0 = u - a * 20s
Simplifying the equation, we find:
a = u / 20
We know that the motor car was moving with uniform velocity for 20 seconds, so the deceleration is equal to the uniform velocity divided by 20 seconds.
Unfortunately, the question does not provide the value of the uniform velocity, so we cannot calculate the deceleration without that information.