Nonuniform Linear Acceleration

[Auburn2017]

Homework Statement

Refer to figure.

Homework Equations

v=ds/dt → ds=vdt
a=dv/dt → dv=adt
ads=vdv

The Attempt at a Solution

I tried taking the derivative of the given position function. Then I am kind of lost from there.

 

[Hamal_Arietis]

First, you must find velocity equation: v=ds/dt=f(t)
After that v=40m/s then f(t)=v=> t0
Finally, find aceleraition equation: a=dv/dt =g(t)
Replace the t0 a(t0)=g(t0)

 

[Auburn2017]

First, you must find velocity equation: v=ds/dt=f(t)
After that v=40m/s then f(t)=v=> t0
Finally, find aceleraition equation: a=dv/dt =g(t)
Replace the t0 a(t0)=g(t0)

If you do this then you don't get the solution that was provided...

 

[Hamal_Arietis]

Ah two time are different. It is the time that you find aceleration
Find the time from v=15 to v=75
From v(t) equation: f(t1)=15m/s => t1
f(t2)=75m/s => t2 So the time is t2-t1

 

[cnh1995]

If you do this then you don't get the solution that was provided...

If initial velocity is the velocity at t=0, it is not 15m/s. If you neglect the 15m/s, your answers will match the given answers.

 

[haruspex]

If initial velocity is the velocity at t=0, it is not 15m/s. If you neglect the 15m/s, your answers will match the given answers.

Seems to me you get the given answers by taking the initial speed as 15m/s and changing the displacement to be 3t³+15t+6 to match.

 

[cnh1995]

Seems to me you get the given answers by taking the initial speed as 15m/s and changing the displacement to be 3t³+15t+6 to match.

Right.. I meant to say neglect 14m/s and take initial velocity as 15m/s. I was about to edit but my network went off and later when I signed up again, I totally forgot that I was going to edit the post. Thanks!