Chikis wrote:A particle moving in a straight line with uniform deceleration has a velocity of 40ms-1 at a point P, 20ms-1 at a point Q and comes to rest a point R where QR=50, calculate the (i) distance PQ; (ii) time taken to cover PQ (iii) time taken to cover PR.
How do about calculating this?
Here is another way of looking at it:
We could plot a velocity-time graph to represent the motion of the particle. We know that the gradient of a line on a velocity-time graph represents the acceleration/deceleration of the particle and since we're told that the particle decelerates uniformly from point P (with velocity of 40ms-1) to point Q (with velocity of 20ms-1) then comes to rest at point R. This means we have a straight line that slopes down.
Given the distance between QR is 50m, but we know we can find the distance between QR by determining the area under velocity-time line, in this case, we have a right triangle here.
Distance of QR
= Area under the line of velocity-time graph from point Q to point R
=Area of the right triangle
Solving the equation above for T, we get T=5s
We know the gradients of the line PQ and QR are the same (from the fact that the particle moves with uniform deceleration), therefore, we have
Hence, this is the answer the second part of the question, that is, the time taken to cover PQ = t = 5s.
And the answer to the third part of the question, time needed to cover PR = (5 + 5)s =10s.
Now, let's go back to the first question.
We're asked to find the distance between the point P and Q.
Distance of PQ
= Area under the line of velocity-time graph
=Area of the trapezium