Please help with this physics question!!! please! im not sure if im doing it right! thanks!

Question

A 1400-kg automobile is driven down a 4 degree incline at a speed of 88 km/h when the brakes are applied, causing a total braking force of 7500 newtons to be applied to the automobile. determine the distance traveled by the automobile before it comes to a stop.

Answer 1

From Newton's 2nd Law of Motion,

F = ma

where

F = braking force = 7500 N

m = mass of the automobile = 1400 kg.

a = acceleration

Substituting appropriate values,

-7500 = (1400)(a)

NOTE the negative sign attached to the force (7500 N). It simply denotes that the direction of the braking force is opposite that of the car's direction.



Solving for "a",

a = - 7500/1400 = - 5.36 m/sec^2

The next formula to use is

Vf^2 - Vo^2 = 2as

where

Vf = final velocity of car = 0 (when the car stops)

Vo = initial velocity = 88 mph = 24.44 m/sec.

a = acceleration (as calculated above) = -5.36 m/sec^2

s = stopping distance

Substituting appropriate values,

0 - 24.44^2 = 2(-5.36)s

Solving for "s",

s = 24.44^2/(2 * 5.36)

s = 55.72 meters

NOTE -- I think driving down an incline (given the particular data in this problem) is not a factor to consider.

Answer 2

a diagram in this case is really helpful

so lets first lets convert units to SI units (metric)

so 88km/h is equal to 24/4/9 m/s

Notice this braking force is acting against the gravitational force (weight); the ANGLE IS VERY IMPORTANT the steeper the slope of incline the harder it is to brake therefore there will be very limited force and deceleration, with a 4 degree incline it will take a longer time to brake then on a flat horizontal surface (good to keep this point in mind)

so we have to calculate the gravitational force that is pulling down on the car parallel to the incline,

F=ma

F=1400*9.8

F= 13720

now parallel to the incline

F=13720*cos (90-4)

F= 957.0588198 N

now calculate the NET braking force of the car

F (net) = 7500- gravitational force

= 7500- 957.0588198

= 6542.94118 N

This net force is acting AGAINST the car since it is slowing down or decelerating the car so it is negative,

- 6542.94118 N

now use Newton's Second Law F=ma to calculate acceleration but in this case deceleration, force is negative since it acts against the car

F=ma

-6542.94118 =1400*a

a= -4.673529414 m/s^2

now calculate time it takes to stop

a=v/t

-4.673529414=(0 - 24/4/9)/t

t= 5.230403465 s

now calculate distance

d= vt+0.5at^2

d= 24/4/9*5.230403465+0.5(-4.673529414)*

(5.230403465)^2

d= 63.92714895 metres

So it takes 63.93 metres to stop the car

hope this helps

Answer 3

Draw a free body diagram:

F(brake) - mg*sin(4)=m*a  where a is the deceleration of the automobile.

a=(vf - v0)/t^2

solve for t.

v0=24.444 m/s.

t=5.23s

a=-.893653 m/s^2.

then distance=v0*t + .5*a*t^2

                   = 115.62 m

Answer 4

acceleration down the slope = gsin(theta) - Ffriction/ mass

=  9.8 sin(4degrees) - 7500/1400 which is negative (slowing down)

vfinal^2 = 0 = Vinitial^2 + 2*acceleration*deltax

vinitial = 88 *1000/ 3600 m/s

solve for deltax