The weight of the vehicle is suspiciously low, being the weight of a man. A vehicle travelling at 100ft/s is travelling a little less than 70 mph. A quick calculation shows the coefficient of kinetic friction µ to be in excess of 1, whereas it should be between 0 (no friction) and less than 1 for a road, even on the driest days. I suggest the vehicle weight is 2000lb and the calculations below use that assumption. 1HP=33000 ft-lb force per minute=550 ft-lb/sec.

The power of the vehicle must be sufficient to overcome the force of friction opposing the motion.

N=normal force of the vehicle’s weight=2000lb force. Frictional force=2000µ lb force.

Power = force×speed = 2000µ×100=200000µ ft-lb force/sec = 50HP = 27500 ft-lb force/sec. So µ=27500/200000=0.1375.

The angle of the slope (1 foot rise for every 100 feet of road) θ is given by sinθ=1/20=0.05. The normal force on the road is diluted by the slope=2000cosθ=1997.5 approx, so the frictional force=1997.5µ=274.656 lb force. The weight of the vehicle also opposes the motion and on the slope this is equivalent to 2000sinθ=100 lb force. The total opposing force is 374.656lb force. The speed is v so 374.656v=27500, and v=27500/374.656=73.4 ft/sec approx.

If the weight of the vehicle is 200lb, µ=1.375 instead of 0.1375. On the slope the frictional force remains the same but the weight of the vehicle pulling down the slope is 10lb instead of 100lb, and the reduced speed is 27500/284.656=96.61 ft/s approx.