a driver shifts into neutral when her 1200 kg is moving at 80 km/h and finds the speed has dropped to 65 km/h 10 s later . what was the average drag force acting on the car?

Question

espex · Answer

Acceleration is constant so list off your known and unknown:
$$m=1200kg, t _{f}=10 \sec, t _{i}=0\sec$$
Want to know drag force $$F_d=?$$
$$v _{i}=\frac{80km}{1hr} \times \frac{1hr}{3600 \sec} \times \frac{1000m}{1km}=\frac{22.2m}{\sec}$$
$$v _{f}=\frac{65km}{1hr} \times \frac{1hr}{3600 \sec} \times \frac{1000m}{1km}=\frac{18.06m}{\sec}$$
$$a=\frac{v _{f}-v _{i}}{t _{f}-t _{i}} \rightarrow \frac{18.06m/s - 22.2m/s}{10\sec - 0\sec}=\frac{-0.414m}{\sec ^{2}}$$
$$F=ma \rightarrow 1200kg \times \frac{-0.414m}{s^2}=-496.8N$$

anonymous · Answer

the average drag force acting on the car will be equals to deacclearation$$\times$$total mass of the car (including driver if any)


now ,

convert all speed in m/sec and use equation 

V=U+AT
find accl and you are done!!

anonymous · Answer

follow up question for this problem; if the driver presses down on the gas pedal and thereby applies a force of 1300 N to the car. How long does it take the car to go from 65 km/hr to 80 km/hr