The height h metres of a ball above the ground at time t secs is h = 5t(4-t) How do you find the average velocities of the ball during the first and third seconds and the average speed? Can you also explain the difference btwn velocity/speed/acceleration?

1 Answer
A08
Jan 24, 2018

Given is expression between height #h# of a ball and time #t# as

#h = 5t(4-t)#
#=>h = 20t-5t^2#

Rewriting (2) as kinematic expression

#h=ut+1/2g t^2# ....(1)
#h = 20t+1/2(-10)t^2# ....(2)

We observe that at #t=0#, #h=0#, initial velocity #u=20ms^-1#, acceleration due to gravity #=-10ms^-2#.

The equation represents a ball thrown upwards from ground level with initial velocity #u=20ms^-1#. #g# has a negative sign in front of it as its direction is opposite to the initial direction of movement of ball. It acts as retardation on the ball.

Kinematic expression for velocity is

#v=u+at#

Velocity after #t=1s#

#v_1=20+(-10)xx1#
#=>v_1=10ms^-1#
Direction of motion is upwards.

Average velocity during first second

#v_(avg)=(u+v_1)/2=(20+10)/2=15ms^-1#
Direction of motion is upwards.

Speed is magnitude of velocity. As such there is no direction associated with speed.
Average speed during first second is #=15ms^-1#.

Velocity after #t=2s# is

#v_2=20+(-10)xx2#
#=>v_2=0ms^-1#
The ball is stationary.

Velocity after #t=3s# is

#v_3=20+(-10)xx3#
#=>v_3=-10ms^-1#
Direction of motion is downwards, as we defined upwards as #+ve#

Average velocity third second

#v_(avg)=(u_2+v_3)/2=(0-10)/2=-5ms^-1#
Direction of motion is downwards

Average speed during third second is #=5ms^-1#.
.-.-.-.-.-.-.

Acceleration is rate of change of velocity with time.
Velocity is rate of change of displacement with time.
Both these are vector quantities.

Speed is magnitude of velocity. As such there is no direction associated with speed. It is a scalar quantity.

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