A car can decelerate at -4.80 m/s^2 without skidding when coming to rest on a level road. What would its deceleration be if the road were inclined at 13 degrees uphill? Assume the same static friction force.

Question

Supposively the answer is -7.00 m/s^2 but I keep on getting -2.47 m/s^2 because I interperted it up the hill and decelerating

nevertheless when I consider it as the car going down hill and decelerating I'm getting -6.88 m/s^2

can you please show me how to get -7.00 m/s^2

Thanks

Answer 1

The deceleration rate is the ratio of applied force (backwards along the direction of motion) to the mass.

The mass remains the same. Therefore the acceleration is higher by the ratio
(Fstatic + M g sin 13)/Fstatic

You also know that the decelerating force applied by the tires is
Fstatic = M*4.80,
(from Newton's law), and that it stays the same

Therefore
(deceleration-uphill)/(deceleration-level)
= 1 + M g sin 13/M*4.80
= 1 + g sin 13/4.80 = 1.4597
(deceleration-uphill) = 1.4597*(-4.80)
= -7.007 m/s^2

If I had used 9.80 instead of 9.81 m/s^2 for g, I would have gotten -7.004 as an answer