A particle starts its motion from point A in the positive x direction in such a way that after every t sec it is at a distance of t2 from the starting point. Find its instantaneous velocity at t=4th sec also the rate at which velocity changes.
v = 8 m/s rate of velocity change = 2
Given at any time the distance of the particle from origin will be equal to t2
In other words x = t2
To find instantaneous velocity we need dxdt
⇒x(t)=t2
dxdt=2t
v = 2t
So at t = 4 sec
v = 2 \times 4 = 8 m/s
Rate at which velocity changes
V2−V1t2−t1
Let's take t1=0
t2=4
V at t2=8m/s
So rate = 8−04−0=2
Also rate can be found by dvdt
V = 2t
So dvdt=2