Answer:
a) F = -1035.385 N
b) Backwards
c) s = 15.60 m
Step-by-step explanation:
Given information
= Initial Speed of Car = 15.0 m/s
= Final Speed of Car = 9.00 m/s
= Breaking Time = 1.30 s
= Mass of Car = 1040 kg
Part (a)
To find the force exerted on the car we use the following formula
![F = ma](https://img.qammunity.org/2020/formulas/physics/high-school/6dslirwz7efajuroewtnbxamr1m7rpia2b.png)
Where
= Force = unknown
= Mass of Car = 1040 kg
= Acceleration of Car / Deceleration of Car = unknown
To find the force (F) we need to first find the deceleration rate (a)
To find the deceleration rate we use the following formula
![a = (v - u)/(t)](https://img.qammunity.org/2020/formulas/physics/middle-school/ipvv7rfojoflxx23js1ndr8jvcpdypbgh9.png)
Inputting the given values
![a = (15 - 9)/(1.30) \\ a = -4.615](https://img.qammunity.org/2020/formulas/physics/high-school/jyu2yp2090y1q3csl126t6118154gklc70.png)
To find the force
![F = ma \\ F = (1040)(-4.615) \\ F = (1040)(-4.615) \\ F = -1035.385 N](https://img.qammunity.org/2020/formulas/physics/high-school/escd4sq7wvdqojpdnth4s07lnykg7a2xz6.png)
Part (b)
Since the value of F is negative this means the the force was opposite the direction of motion, hence the force was backwards.
Part (c)
To find the total distance the car moved while braking we use the following formula
![v^2 = u^2 + 2as](https://img.qammunity.org/2020/formulas/physics/high-school/a36drfxj6flq91dulkeciotm4pniqz3c07.png)
Where
= distance traveled
Inputting the values given
![(9)^2 = (15)^2 + 2(-4.615)s](https://img.qammunity.org/2020/formulas/physics/high-school/8ebdv2qn6c4pfqil6yqgo31gctjw4zu3uu.png)