A car moves with a constant velocity of 10m/s for 10s along a straight track then it moves with uniform acceleration of 2m/s^2 for 5 seconds. Find the total displacement and velocity at the end of the 5th second of acceleration.
Answers
Answer:
Firstly we look at this rather straightforward and classic kinematics problem in two parts ….
The first part of the motion of the car in discussion is a uniform type of motion where there is no variation in the acceleration with time .
Therefore for the specified time interval i.e. for 10 seconds the car moves with a constant velocity of 10 m/s and is displaced by 10 m/s x 10s = 100 m (Assuming the car is moving in a straight line .)
Therefore displacement at the end of 10 s = 100m
The second part of the problem deals with the body being accelerated .
Therefore as a consequence of this uniform acceleration ( the acceleration does not vary with time i.e =2m/s2) , the velocity will increase linearly with time given by the equation v= u + at
the parameters are as follows : u= initial velocity=10m/s
a= acceleration= 2m/s2
t= time interval under consideration
thus the final velocity at the end of the 5 seconds of accelerated motion is
10 m/s + (2m/s2)(5s)= 20 m/s
Therefore the displacement at the end of the 5 seconds is given by
s= ut +1/2 at2
which when the values are substituted should give you an answer of 75 m
BUT……….
we have totally forgotten about the displacement in the initial 10 seconds of the motion .
Therefore the total displacement at the end of 15 seconds is 100m +75m= 175m
Hope this helps….
Answer:
For the current problem, it is good ignore that the car is moving for 10s at 10m/s.
Initial Velocity 10m/s, Acceleration 2m/s/s.
Hence, Distance Covered During First 5 seconds of Acceleration = (Initial Velocity)x(Time) + (Acceleration)x(Time)x(Time)/2
= (10)x(5)m + (2)x(5)x(5)/2 m = 50m + 25 m = 75m
Velocity at end of 5th second = (Initial Velocity) + (Acceleration)x(Time)
= 10m/s + 2x5 m/s = 20m/s
Total Distance Covered = Distance Covered During First 10 seconds + Distance Covered During First 5 seconds of Acceleration
= 10x10m + 75m = 100 + 75m = 175m