A car can decelerate at -3.15 m/s2 without skidding when coming to rest on a level road. What would its deceleration be if the road is inclined at 9.7∘ and the car moves uphill? Assume the same static friction coefficient.

Deceleration = 5.7 m/s²

Explanation:

We are given;

Deceleration of the car; d = -3.15 m/s²

Angle of inclination; θ = 9.7°

Coefficient of friction will be calculated as; μ = d/g = 3.15/9.81 = 0.3211

Now,the acceleration would be gotten from the formula;

a = g (sin(θ) + μ cos(θ))

Plugging in the relevant values to give;

a = 9.81[sin 9.7 + (0.3211 × cos (9.7)]

sin 9.7 and cos 9.7 are in radians, thus we have;

a = 9.81(-0.2718 + (0.3211 × -0.9624))

a = -5.7 m/s²

Negative acceleration means a Deceleration of 5.7 m/s²

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//