CALC? A car’s velocity as a function of time is given by vx(t) = ? + ?t2, where ? = 3.00 m/s and ? = 0.100 m/s3. (a) Calculate the average acceleration for the time interval t = 0 to t = 5.00 s. (b) Calculate the instantaneous acceleration for t = 0 and t = 5.00 s. (c) Draw vx-t and ax-t graphs for the car’s motion between t = 0 and t = 5.00 s.

Solution 17E The velocity as a function of time is given as, 2 V (t) = + t --------------------(1) When = 3 m/sand = 0.100 m/s , 3 a) Average acceleration for the time interval of t=0 to t= 5s is, V (5)V (00 a = 50 Now, V (5) = 3 + 0.100 × 5 = 3 + 0.100 × 25 = 3 + 2.5 = 5.5 m/s 2 V (0) = 3 + 0.100 × 0 = 3 m/s. a = V (5)V (0= 5.53= 2.5 = 0.5 m/s . 50 50 5 b) As we know acceleration is the time derivative of velocity, a = dV = d( + t ) = 2t. dt dt This is the instantaneous acceleration as a function of time. So, at t = 0, the instantaneous acceleration will be, a(0) = 2t = 2 × 0.100 × 0 = 0 m/s .2 And at t = 5, the instantaneous acceleration will be, a(5) = 2t = 2 × 0.100 × 5 = 1 m/s . 2