Calculating Braking Force and Stopping Distance for an Automobile

[Sneakatone]

With the brakes fully applied , a 1470 kg automobile decelerates at the rate of 7.5 m/s^2.
a) what is the magnitude of the breaking force acting on the automobile?
1470 *7.5=11025N ( I think this is correct)

b)if the initial speed is 90km/h (25 m/s) what is the stopping distance?
im thinking 25/7.5=3.33s
3.33*25=83.25 m

c)what is the work done by the breaking force at 90km/h.
part a) *part b)=J

d) what is the change in kinetic energy of the automobile?
1/2mv^2

I believe I know how to do the proceeding parts but I need the previous values.

 

[Doc Al]

a) what is the magnitude of the breaking force acting on the automobile?
1470 *7.5=11025N ( I think this is correct)

OK.

b)if the initial speed is 90km/h (25 m/s) what is the stopping distance?
im thinking 25/7.5=3.33s
3.33*25=83.25 m

Careful: The speed is not constant.

 

[bossman27]

(b) is incorrect. The distance formula is d = v_{i}t+\frac{1}{2}at^{2}. Note that acceleration is negative in this case, and see if you can go from there.

 

[Sneakatone]

if I use d = vt + (1/2)at^2
d=(25)(3.33)+1/2(-7.5)(3.33)^2
d=41.66
did I use the correct acceleration?

 

[Sneakatone]

my distance is correct but for part c when I multiplied 41.66 * 11025N= 459301.5 J it is wrong.

 

[Sneakatone]

never mind the work is negative, Thanks for the help!