Q: Before picking up the passengers, the driver of the car of Exercise 49* shift
ID: 1284947 • Letter: Q
Question
Q: Before picking up the passengers, the driver of the car of Exercise 49* shifts into neutral when the car is moving at 80 km/h and finds that its speed has dropped to 65 km/h after 10s. What was the average drag force?
* Q49: A car and the driver with a total mass of 1600 kg has a maximum acceleration of 1.2 m/s^2. If the car picks up three 80-kg passengers, what is the maximum acceleration now?
Chegg says that the answer is v2-v1= 4.17m/s; F=ma=m(v2-v1)/t = 500N
My calculations are F= ma = 1600 (4.17)/10 = 667.2N .... So either the 500N is wrong, or I am seriously confused on something. Help!
Explanation / Answer
v1 = 80 km/h = 80*5/18 = 22.2 m/s
v2 = 65 km/h = 65*5/18 = 18.06 m/s
a = (v2-v1)/t
= (18.06-22.2)/10
= -0.414 m/s^2
drag Force, F = m*a
= 1600*0.414
= 662.4 N <<<<<<<<------Answer
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