A car is traveling along a road, and its engine is turning over with an angular velocity of +190 rad/s. The driver steps on the accelerator, and in a time of 13.0 s the angular velocity increases to +300 rad/s. (a) What would have been the angular displacement of the engine if its angular velocity had remained constant at the initial value of +190 rad/s during the entire 13.0-s interval? (b) What would have been the angular displacement if the angular velocity had been equal to its final value of +300 rad/s during the entire 13.0-s interval? (c) Determine the actual value of the angular displacement during the 13.0-s interval.

asked by Sanz

10 years ago

659 views

0

0

1 answer

a. D = Vo*t = 190rad/s * 13s = 2470 rad

b. D = 300rad/s * 13s = 3900 Rad.

c. a = (V-Vo)/t = (300-190)/13 = 8.46 rad/s^2.

D = Vo*t + 0.5a*t^2
D = 190*13 + 4.23*13^2 = 3185 Rad.

answered by Henry

0

0

Ask a New Question or answer this question.

Question

A car is traveling along a road, and its engine is turning over with an angular velocity of +190 rad/s. The driver steps on the accelerator, and in a time of 13.0 s the angular velocity increases to +300 rad/s. (a) What would have been the angular displacement of the engine if its angular velocity had remained constant at the initial value of +190 rad/s during the entire 13.0-s interval? (b) What would have been the angular displacement if the angular velocity had been equal to its final value of +300 rad/s during the entire 13.0-s interval? (c) Determine the actual value of the angular displacement during the 13.0-s interval.

1 answer

a. D = Vo*t = 190rad/s * 13s = 2470 rad

b. D = 300rad/s * 13s = 3900 Rad.

c. a = (V-Vo)/t = (300-190)/13 = 8.46 rad/s^2.

D = Vo*t + 0.5a*t^2
D = 190*13 + 4.23*13^2 = 3185 Rad.

Ask a New Question or answer this question.

Henry · Answer

a. D = Vo*t = 190rad/s * 13s = 2470 rad b. D = 300rad/s * 13s = 3900 Rad. c. a = (V-Vo)/t = (300-190)/13 = 8.46 rad/s^2. D = Vo*t + 0.5a*t^2 D = 190*13 + 4.23*13^2 = 3185 Rad.