Derivatives?

Question

If v = t^2 + 4m/s what is the acceleration at times (acceleration not constant):
t=1s and t=4s

I've worked out that V=5 at 1s and V=20 at 4s using the initial formula...

And then supposedly meant to use this formula to find accelleration => a=dv/dt...

But if you know an easier way that would be even better

Thanks guys

Yeah sorry... i dont know how to use that formula => how do i derive using dv/dt? what is dv?

Yeah that helps - Thanks... But i still dont understand why a = 2t?

Answer 1

a=dv/dt
but you knew that

a(t)= 2t yes it seems to change with time.
a(1)= 2 m/s^2

a(4) = 8m/s^2

also for average acceleration

a(avg) =(V2-V1)/(t2 - t1)= (20-5)/3=5m/s^2

curious coincidence

a(avg)= [a(4)+ a(1)]/2=[8 + 2]/2= 5 m/s^2

Okay
Do you know Calculus?
v(t)=ds/dt
a(t)=dv/dt=d^2s/dt^2 So this does not help?

By definition
The derivative of a function is defined as a limit of t2-t1 approaching 0 of v(t2) -v(t1)/ (t2-t1) . The thus it is a slope at a point on that function.

Answer 2

dv/dt=2t + c thats acceleration.
when t is 4 seconds then acceleration is 8meters per second squared