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June 2015 p42 q5
3834
A particle P starts from rest at a point O on a horizontal straight line. P moves along the line with constant acceleration and reaches a point A on the line with a speed of 30 m s-1. At the instant that P leaves O, a particle Q is projected vertically upwards from the point A with a speed of 20 m s-1. Subsequently P and Q collide at A. Find
the acceleration of P,
the distance OA.
Solution
(i) To find the acceleration of P, we first determine the time taken for Q to return to point A. Using the equation for vertical motion,
\(-20 = 20 - 10t\)
Solving for \(t\), we get \(t = 4\) seconds.
Now, using the equation \(v = u + at\) for P, where \(v = 30\) m/s, \(u = 0\), and \(t = 4\) s,
\(30 = 0 + 4a\)
Solving for \(a\), we find \(a = 7.5\) m/s2.
(ii) To find the distance OA, we use the equation \(v^2 = u^2 + 2as\), where \(v = 30\) m/s, \(u = 0\), and \(a = 7.5\) m/s2.