[cdedmundson]

I know I will need to use some integration but I am not yet familiar enough with it to use it successfully. It would be greatly appreciated if someone could walk me through this problem...

The acceleration of a bus is given by the equation a(t)=βt where β=1.2m/s^2.

a.) if the bus's velocity at time t=1.0s is 5m/s, what is the bus's velocity at 2s?
b.) 8.7s?

[stingRay18]

I know I will need to use some integration but I am not yet familiar enough with it to use it successfully. It would be greatly appreciated if someone could walk me through this problem...

The acceleration of a bus is given by the equation a(t)=βt where β=1.2m/s^2.

a.) if the bus's velocity at time t=1.0s is 5m/s, what is the bus's velocity at 2s?
b.) 8.7s?

Given V(t) = Vi + a*t where Vi = Initial Velocity.

If V(1) at 1 sec = 5 Then solving for Vi
5 = Vi + (1.2*t)*t
5 = Vi + 1.2*(t^2)
5 = Vi + 1.2(1^2)
5 = Vi + 1.2
Vi = 3.8

Then solving for V at 2 and 8.7 sec
V(2) = 3.8 + 1.2*(2^2) = 8.6
V(8.7) = 3.8 + 1.2*(8.7^2) = 94.628

That's how I see it.

[cdedmundson]

that would work if the acceleration was CONSTANT, but since the acceleration is changing with relation to time you need to use integration...

I figured this out about 20 minutes after I posted this problem and anyone is welcome to use this example to learn basic integration...

Integration is also known in some other countries as "anti-derivatives", which is exactly what they are. The integral equation to solve for the acceleration is ∫(0→t) β/2*t^2 + c, you solve for c at t=1 and you get c=4.4, plug c back into the equation and solve for the total equation at 2 and 8.7 (6.8 and 49.8 respectively)

[stingRay18]

That makes sense - thanks.

I need to take the rust off my calculus :)

Be well.