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# M2 - Basic Integration of Velocity question watch

1. Yeah I can't seem to answer this question correctly, I'm told that a particle P is moving along the x-axis, the velocity of P at a given time 't' is given by ((3t^2)-12t+5) in the direction of x-increasing. At t=0, P is at the origin.

I'm told to work out the velocity when the acceleration is zero, 7m/s in the direction of x decreasing, and to find 't' when P is at the origin again which is 1 and 5s. I can do that but the question is I have to find the distance travelled in the interval 3s to 4s (where t is greater of equal to 3 and less than or equal to 4)

The displacement from integrating the velocity equation tells me that they are both displaced the same amount from the origin, so I have no idea about how to go about this.

Thanks Guys.
2. draw the graph of the velocity. area under graph should give you the displacament
3. Yeah I tried that but the graph would be a quadratic and then to work out the area under the graph you'd use integration, which hasn't given me the correct answer of 48m. I'm not saying that the method of integration is incorrect, I think I've made a mistake somewhere, for a velocity ((3t^2)-12t+5) at at t=0 it's at the origin therefore constant of integration is zero, would the integral be ((t^3)-(6t^2)+5t)?

Thanks.
4. between 3 and 4, the graph cuts the x-axis. so one part of the graph will give you a +ve area and one part will give you a negative area. you have to add both the parts up, irrespcetive of their signs
5. yes your integral is correct
6. Thanks Kosy91! I've done it now.
7. oh...did you get it??? how did you do it??

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