The position of a particle moving along an x axis is given by x = 12.0t2 - 6.00t3, 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 = 7.00 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)? (i) Determine the average velocity of the particle between t = 0 and t = 7.00 s.
Explanation / Answer
Given the position of the ball is given by x = 12.0 + 6.0t^3 Imagine the body starts from origin and travelling on x-axis. at time t = 2s, x2 = 12*4+6*8=98 cm at time t = 3s, x3 = 12*9+6*27= 270cm Average velocity is given by displacement / time taken. We know from above that the position of the ball is 98cm at 2sec and it is 270cm at 3 sec along the same line (x-axis). Hence the resultant displacement = delta x = x3 - x2 =172 cm and the time taken = 3 - 2 = 1 sec. Hence the average velocity = 172cm/sec