Question

17. A distracted driver going along a parkway at constant speed of 14 m/s passes a police car parked at the sidelines. If the police car takes 4.0 s to take off and chases the distracted driver accelerating at a rate of 2.0 m/s2, how long until it catches with the distracted driver's car? A. 28 s B. 21 s C. 7.0 s D. 110 s E. 56 s

Answer 1

Given details are: speed (v) = 14 m/s time (t) = 4 s acceleration (a) = 2 m/s^2 The police take 4s to take off, while the driver passes him. so in this span, driver will be at s_dr = v_t + 0.5 a_d t^2 rightarrow (1) = (14 times 4) times (0.5) times (0) times (4)^2 (Because a_dr = 0) s_dr = 56 m Now, we can say that distance covered by the police must be equal to distance covered by the driver plus 56 m Therefore S_p = S_dr + S_d = 56 + S_d using equ (1) we get U_p t + (0.5) a_p t^2 = U_d t + 56 (Because a_dr = 0) (0.5) (2) times t^2 = 14 times t + 56 t^2 - 14 t - 56 = 0 by solving this using quadratic formula, we get t = 14 plusminus squareroot (-14)^2 times 4 times (1) (56)/2 times 1 = 14