The position of a ball rolling in a straight line is given by x= 2.0 - 3.6t + 1.

Question

The position of a ball rolling in a straight line is given by x= 2.0 - 3.6t + 1.1t2, where x is in meters and t in seconds. (a) determine the position of the ball at t = 1.0s, 2.0s, and 3.0s. (b) What is the average velocity over the interval t = 1.0s to t = 3.0s? (c) What is its instantaneous velocity at t = 2.0 s and at t = 3.0 s?

Must show fundamentals to get formulas for instantaneous velocity. Like proof with a limit of how you know that the instantaneous velocity is the derivative of the position function. This is where I am stuck. This is problem 16 from Chapter 2 in Giancoli's Physics for Scientists and Engineers with Modern Physics 4th edition. There was not an answer posted.

Explanation / Answer

Hi, Given the position of the ball is given by x = 2 - 3.6t + 1.1t^2 Imagine the body starts from origin and travelling on x-axis. a) at time t = 1s, x1 = 2 - 3.6 + 1.1 = -0.5 m (-ve sign implies for -ve x-axis direction)     at time t = 2s, x2 = 2 - (3.6 * 2) + (1.1 * 4) = 2 - 7.2 + 4.4 = -0.8 m     at time t = 3s, x3 = 2 - (3.6 * 3) + (1.1 * 9) = 2 - 10.8 + 9.9 = +1.1 m b) Average velocity is given by displacement / time taken.     We know from above that the position of the ball is -0.5m at 1sec     and it is +1.1m at 3 sec along the same line (x-axis).     Hence the resultant displacement = delta x = x3 - x1 = (1.1) - (-0.5) = 1.6m         and the time taken = 3 - 1 = 2 sec.    Hence the average velocity = 1.6 / 2 = 0.8m/sec c) Now we know the expression for x at a time t is given by x = 2 - 3.6t + 1.1t^2     so let us imagine the particle moves a distance dx in a time dt.     then this distance can be found by finding derivative of above equation given by       dx = 0 - 3.6dt + 1.1*2tdt = (2.2t - 3.6)dt.     And we know the expression for velocity is given by v = dx/dt     Hence expression for instantaneous velocity v in this case = (2.2t - 3.6)dt.    Now the instantaneous velocity at time t = 2sec = v2 = (2.2 * 2) - 3.6 = 0.8m/s and it will be along +ve x-axis (since the value is +ve).     And the instantaneous velocity at time t = 3sec = v3 = (2.2 * 3) - 3.6 = 3m/s and it will be along +ve x-axis (since the value is +ve). Hope this helps you. Hope this helps you.

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