DOUBT IN INSTANTANEOUS ACCELERATION AND VELOCITY

Question

i have understood the definitions of both instantaneous acceleration and velocity. i am also clear about average acceleration and velocity. Instantaneous acceleration is the acceleration of the object at some instant in time and can be found by taking the derivate of the velocity function.

 but i still have this confusion around instantaneous acceleration, how exactly does instantaneous acceleration affects the velocity.

For eg; lets take this problem.

A particle is in motion and is accelerating. The functional form of the velocity is v(t) = 20t − 5t^2 m/s.

Find the instantaneous velocity at t = 1, 2, 3, and 5 s.

Find the instantaneous acceleration at t = 1, 2, 3, and 5 s.

Solution

a=dv/dt= 20-10t m/s^2

v(1 s) = 15 m/s, a(1 s) = 10m/s^2

v(2 s) = 20 m/s, a(2 s) = 0m/s^2,

v(3 s) = 15 m/s, a(3 s) = −10m/s^2,

v(4 s) = 0 m/s, a(4 s) = -20m/s^2

v(5 s) = −25 m/s, a(5 s) = −30m/s^2

Now lets see instantaneous acceleration and velocity between 1 to 2 sec

Avg acceleration between 1 to 2 sec would be v2-v1 /t2-t1 = 5m/s^2.

v(1 s) = 15m/s ,   a(1 s) = 10m/s^2

v(1.1s) = 15.95m/s, a(1.1 s) = 9m/s^2

v(1.2s) = 16.8m/s,  a(1.2 s) = 8m/s^2

v(1.3s) = 17.55m/s,  a(1.3 s) = 7m/s^2

v(1.4s) = 18.2m/s  a(1.4 s) = 6m/s^2

v(1.5s) = 18.75m/s  a(1.5 s) = 5m/s^2

v(1.6s) = 19.2m/s  a(1.6 s) = 4m/s^2

v(1.7s) = 19.55m/s  a(1.7 s) = 3m/s^2

v(1.8s) = 19.8m/s  a(1.8 s) = 2m/s^2

v(1.9s) = 19.95m/s  a(1.9 s) = 1m/s^2

v(2 s) = 20m/s    a(2 s) =  0m/s^2

i just want to know this.

Acceleration is change in velocity right? when acceleration happens at an instant, say v1 might be changing to v2. so the instantaneous change at t=1 is 10m/s^2 i.e at t=1 there was increase in velocity by 10m/s^2? why cant wee see that change in the table? i maybe sounding silly or dumb but i dont know what s happening here.i hope you get what i am asking. thank you.  

Answer 1

At t = 1 sec, the velocity doesn't instantly change by 10 m/s.

Consider this: At t = 1.1 sec, the formula says the velocity is 15.95 m/s. the average acceleration from 15.0 m/s to 15.95 m/s is 0.95 / 0.1 = 9.5 m/s^-2. This doesn't mean that the velocity changed by 9.5 m/s during that 0.1 s. The change in velocity would be 9.5 m/s^-2 x 0.1 s or 0.95 m/s which is what the formula predicts. The acceleration you calculate from the equation 20 - 10t is the instantaneous rate of change of velocity. hope this helps.

Answer 2

Average acceleration is just an average rate of change of velocity just as you said.

Instantaneous acceleration is given by the derivative of the velocity function at a given time.

It does not mean that the instantaneous velocity has to match the average velocity (usually not the case, you can visualize it on a velocity versus time graph), but you can see that at some time the instantaneous velocity takes the value of the average velocity between two points.

You correctly determined a(1) = 10m/s from the function you determined by the derivative.

Also a(2) = 0 using the same function. Although the acceleration keeps changing there is some point where the acceleration is equal to 5 but not for the whole interval.