Acceleration problem. I , please.

[belltos]

1. A motorcyclist who is moving along an x-axis directed towards the east has an acceleration given by a = 6.1 -1.2t m/s2 for t between 0 and 6s. At t=0, the velocity and position of the cyclist are 2.7 m/s, and 7.3m. a) What is the maximum speed achieved by the cyclist? b) What total distance does the cyclist travel between t=0 and t=6 s.

I don't know how to start this question. I assumed I would use the equation x = x0 + v0t +1/2at^2.
v0 = 2.7
x0 = 7.3
a = 6.1 - 1.2t

I tried subbing those quantities into the equation, and then plug in 6 for time, but my answers don't make sense.
I also tried integrating a to get my velocity function and then solve that indefinite integral using 0 and 6 as my boundries... but I'm stuck.. please help

 

[kuruman]

Hi belltos, welcome to PF. The kinematic equation that you have is valid for constant acceleration. Here is the acceleration is not constant. You are correct in integrating to get the velocity function. If you show us exactly what you did, we might be able to get you unstuck.

 

[belltos]

Sorry, I thought I did show what I thought the relevant equations were, and even said what I have tried... either way... I'm thinking I should use the formula x = x0 + v0t + 1/2at^2
with: a = 6.1 - 1.2t
x0 = 7.3m
v0 = 2.7m/s
plugging these into find x = 7.3 + 2.7t + .5(6.1 - 1.2t)t^2
then subbing in t=6, for x position at t=6 gives.. x = 7.3 + 2.7(6) + .5(6.1-1.2(6))6^2
x = 3.7
I don't think this number makes any sense. I was going to use this x value to determine the max. velocity and then x-x0 for the total distance travelled.

Any thoughts?

 

[belltos]

Integrating the acceleration function (6.1 - 1.2t) gives 6.1t - 0.6t^2, and using t = 0,6 as borders I get an answer of 15. I'm not sure where this gets me. I'm thinking it is the max. speed reached by the cyclist.

I could also set 6.1 - 1.2t = 0 and solve for t, which is t = 5.083, which could be the time when the cyclist reaches the max. speed. therefore I'd plug that number into the integrated acceleration function giving - 15.504. which also could be the max. speed...

I'm clearly confused.

 

[kuruman]

Don't use 0 and t as "borders" (limits of integration is the term). When you do this you get a number not a function of time.

Maximum speed is presumably achieved at some intermediate time between 0 and 6 s. To find the velocity as a function of time, use

\int^{v}_{2.7}dv=\int^{t}_{0}(6.1-1.2t)dt

to get v(t). Then calculate v(5.083 s) to get the maximum value. The second part you can easily get if you know v(t). Here you integrate between 0 and 6 s for the time integral and x0 to x for the position integral.

 

[RoyalCat]

Integrating the acceleration function (6.1 - 1.2t) gives 6.1t - 0.6t^2, and using t = 0,6 as borders I get an answer of 15. I'm not sure where this gets me. I'm thinking it is the max. speed reached by the cyclist.

I could also set 6.1 - 1.2t = 0 and solve for t, which is t = 5.083, which could be the time when the cyclist reaches the max. speed. therefore I'd plug that number into the integrated acceleration function giving - 15.504. which also could be the max. speed...

I'm clearly confused.

You forgot your initial values. :)
For t=0, you find that the velocity is 0, if we use your equation. Does that make sense in light of the data?

 

[belltos]

Ahhh... thanks for the help all. Finally figured it out, need to use my initial velocity as the constant after integrating, then v(0) would equal 2.7 and not 0