Question

Problem 3-7 The acceleration of a vehicle takes the form du dt (i3.6-0.06u where u is the vehicle speed in fsec. The vehicle is traveling at 45 f/sec at time To. (a) Determine the distance traveled by the vehicle when accelerated to a speed of 55 ft/sec. (b) Determine the time at which the vehicle attains the speed of 55 ft/sec. (c) Determine the acceleration of the vehicle after 3 sec.

Answer 1

(3.6 - 0.06 v) = [- 0.06 t/2.303] for t = 03 sec 3.6 - 0.06 v = 0.835 v = 46.07 f t/sec therefore acceleration a = V = u + at 46.07 = 45 + a times 3 a = 0.359 f t/sec^2 \r\n solution: d v/d t = 3.6 - 0.06 v v = vehicle speed (H/s) t_0 = vehicle speed = 45 f t/s (2) Distance travelled by vehicle when accelerated to a speed of 55 f t/sec a = dv/d t = 3.6 - 0.06 v for v = 55 f t/s a = 3.6 - 0.06 times 55 = 0.3 f t/sec^2 distance travelled s = v^2 = u^2 + 2 as s = v^2 - u^2/2 a = (55)^2 - (45)^2/2 times 0.03 not equal to 1.6666.67 f t/s (b) determine the time of where vehicle attained a speed of 55 f t/sec S = u t + 1/2 at^2 16666.67 = 45 times t + 1/2 times 0.03 t^2 t^2 + 3000 t - 111111.33 = 0 t = 333.33 sec (e) determine the acceleration of the vehicle offer 3 s d v/d t = 3.6 - 0.06 v