CALC? A remote-controlled car is moving in a vacant parking lot. The velocity of the car as a function of time is given by (a) What are ax(t) and ay(t), the x- and y-components of the car’s velocity as functions of time? (b) What are the magnitude and direction of the car’s velocity at t = 8.00 s? (b) What are the magnitude and direction of the car’s acceleration at t = 8.00 s?

Solution 8E Step 1: Given v = [5 m/s - (0.018 m/s ) t ] i + [2 m/s + (0.55 m/s ) t ] j ˆ From the above equation 2 v x 5 - 0.018t m/s v y 2 + 0.55t m/s Problem (a) Step 1: To find a xnd a y , differentiate vx and v y a = -0.018*2t x a = -0.036t m/s 2 x 2 a y 0.55 m/s Problem (b) Magnitude and direction of Velocity at t = 8s Step 1: v x 5 - 0.018t = 5 - 0.018*8*8 v x 3.848 m/s v = 2 + 0.55t = 2 + 0.55*8 y v y 6.4 m/s Step 2: Magnitude of velocity (v) = v x2+ v y2 2 2 = 3.848 + 6.4 v = 7.47 m/s Step 3: v y Direction of velocity = arctan( v x) = arctan( 6.4 ) 3.848 Direction of velocity = 58.98° from North from East Problem (b) Magnitude and direction of Acceleration at t = 8s Step 1: a x -0.036t m/s 2 a x -0.036*8 a x -0.288 m/s 2 a y 0.55 m/s 2 Step 2: 2 2 Magnitude of Acceleration (a) = a x + a y 2 2 = ( 0.288) + 0.55 a = 0.621 m/s 2