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Physics 67 Online
OpenStudy (anonymous):

I don't like this question. The position of a particle moving along an x axis is given by x = 15.0t^2 - 3.00t^3, where x is in meters and t is in seconds. Determine (a) the position, (b) the velocity, and (c) the acceleration of the particle at t = 5.00 s. (d) What is the maximum positive coordinate reached by the particle and (e) at what time is it reached? (f) What is the maximum positive velocity reached by the particle and (g) at what time is it reached? (h) What is the acceleration of the particle at the instant the particle is not moving (other than at t = 0)?

OpenStudy (anonymous):

And then (i) Determine the average velocity of the particle between t=0 and t=5.00s.

OpenStudy (anonymous):

The velocity of the particle is equal to the rate of change of x with respect to t. That's the gradient of the x-t graph or function. So you need to differentiate x wrt to t. So start with that. If you differentiate the velocity, you will get the acceleration. That should start you off.

OpenStudy (anonymous):

Physics and any sort of math is not my specialty... could you say more?

OpenStudy (anonymous):

Surely. The function you have relates x (the location of the particle) to t (the time at which you want to know where it is). This function is: x=15t^2 - 3t^3 .... plug in 5 for t = answer a This is just a graph we could draw (if we knew its shape) with x as the vertical axis and t as the horizontal. At any given point of this graph the gradient of the graph represents how x (the displacement of the particle) changes with t. (If you think of a straight line graph, x = m.t +c, m is the gradient of the straight line. Fortunately we do not need to knwo the shape of the graph and draw it because we have differentiation to hand. We differentiate the function with respect to t. The differential is written as dx/dt and the rate of change of displacement with time is also the same as the velocity (in the x direction). So in this case: dx/dt = 30t - 9t^2 = velocity at any time t ... plug in 5 = answer b So this is the gradient at any point t in time. Just plug in the t to get the velocity. Similarly, the rate of change of velocity is acceleration. So differentiate again and you get d^2x/dt^2 = 30 - 18t ... plug in 5 = answer c (I am not going too explain how the differentiation works because that is about a fortnight's work.) So, does that ring any bells at all?

OpenStudy (anonymous):

The other answers are just about testing the above relationships. eg the last one. Get t for when the velocity 30t-9t^2=0... t*(30-9t)=0... t=0, as the question says, or 30/9 and so on.

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