Applied Mathematics

Uniformly Accelerated motion

 1997 Q. 1 (a) A particle, moving in a straight line, accelerates uniformly from rest to a speed v m/s. It continues at this constant speed for a time and then decelerates uniformly to rest, the magnitude of the deceleration being twice that of the acceleration. The distance travelled while accelerating is 6m. The total distance travelled is 30m and the total time taken is 6s. Draw a speed-time graph and hence or otherwise, find the value of v. Calculate the distance travelled at v m/s. Solution: For the acceleration: u = 0m/s s = 6m v = vm/s v2 = u2 + 2as For the deceleration: u = vm/s v = 0 and a = 1997 Q.1(b) Two points p and q are a distance d apart. A particle starts from p and moves towards q with initial velocity 2u and uniform acceleration f. A second particle starts at the same time from q and moves towards p with initial velocity 3u and uniform deceleration f. Prove that the particles collide after d/5u seconds if the particles collide before the second particle comes to instantaneous rest, then fd < 15u2 if fd = 30u2 then the second particle has returned to q before collision. Solution: (i) Sp = s = 2ut + 1 ft2 Sq = d - s = 3ut - 1 ft2 Adding these gives d = 5ut (ii) vq = 3u - ft vq > 0 Þ 3u - ft > 0 (iii) Sq = 0 3ut - 1 ft2 = 0 Þ 6u = ft