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Question

Traveling at a speed of 15.2 m/s, the driver of an automobile suddenly locks the wheels by slamming on the brakes. The coefficient of kinetic friction between the tires and the road is 0.670. What is the speed of the automobile after 1.60 s have elapsed? Ignore the effects of air resistance.

Answer 1

The friction F is equal to the coefficient of kinetic friction,u times the Normal force,N. The Normal force is basically the weight of the object,(w) against gravity(g) perpendicular to the plane.

I am assuming the car is on a level plane and not inclined on say a hill. If it was we would have to figure in the angle with which the Normal force is applied in relations to the gravitational downward force.

So

F = u x N = 0.67 x N

N = mass x gravity = w x g = w x 9.8 m/sec^2

and since the decelerational force,D is just the frictional force not including air resistance then F = D. And decelerational force is the mass(w) x acceleration (deceleration in this case, a). Then from the deceleration we can find out how fast it is traveling after 1.6 s

so a = D / w = u x N / w =u x g x w / w = u x g = 0.67 * 9.8 m/sec^2 = 6.566 m/sec^2

so the car decelerates at a rate of 6.566 meters per sec^2. Therefore in 1.6 secs, the car reduced it speed by 1.6x6.566 = 10.51 meters / sec.

Thus the car was traveling at 15.2m/s - 10.5m/s = 4.7 m/sec.

FInal answer is 4.7 meters /sec

Answer 2

The acceleration is -0.67*g in the direction of travel and constant for 1.6 seconds,so the reduction in speed is 1.6*g*0.67 m/s (if your g is in m/s/s).

Best of luck - Mike

Answer 3

Fµ=Fnet
Fnet=ma
µmg=ma
µg=a

vf=vi+at
vf=vi+µgt
vt=(15.2 m/s)+(.670)(-9.81 m/s²)(1.60 s)
vt=4.68 m/s

Answer 4

x(t) = 0.5 a t^2 + v0 t + x0
a = 0.67 * g = -6.566, v0 = 15.2, x0 = 0, t = 1.6
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x(1.6) = 7.5 m (rounded down)

Oh, v(t) = a*t + v0 = 4.7 m/s (rounded up)