- #1

carodog

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- Homework Statement
- A student is attempting to push his stalled car out of an intersection with his girlfriend at the wheel. The car and girlfriend has a combined mass of m = 965 kg. Unfortunately the hill has an incline of θ = 2.2 degrees with respect to the horizontal. The student can supply a force of F = 852 N for t = 13 s before tiring. What is the furthest the edge of the intersection can be (d) in meters from the stalled car in order to make it out in one push? Assume no rolling resistance from the car and that his girlfriend brakes to bring the car to a stop when he stops pushing.

- Relevant Equations
- m = 965 kg

θ = 2.2 degrees

F = 852 N

t = 13 s

Hints given:

-Start with free body diagram. Use the relationship between impulse and momentum to find the final velocity of the car after he has pushed for time

-Use a kinematic equation to relate the final velocity and time to the distance traveled.

-What is his initial velocity?

My attempt:

Impulse = F (delta t) = change in momentum = 852N (13s) = 11076 Ns

Initial momentum = 0 Ns

Final momentum = 11076 Ns = mv = (965kg) v

Final velocity v = 11.48 m/s

Not sure where to go from here. Thinking maybe I should be splitting things up into x- and y- components and then using the v-t relation: delta v = a (delta t). If initial velocity is 0 (braking) then I could solve for acceleration and then use x-t relation: (delta x) = (initial v)(delta t) + (.5a(delta t)^2) to find displacement...

-Start with free body diagram. Use the relationship between impulse and momentum to find the final velocity of the car after he has pushed for time

*t*.-Use a kinematic equation to relate the final velocity and time to the distance traveled.

-What is his initial velocity?

My attempt:

Impulse = F (delta t) = change in momentum = 852N (13s) = 11076 Ns

Initial momentum = 0 Ns

Final momentum = 11076 Ns = mv = (965kg) v

Final velocity v = 11.48 m/s

Not sure where to go from here. Thinking maybe I should be splitting things up into x- and y- components and then using the v-t relation: delta v = a (delta t). If initial velocity is 0 (braking) then I could solve for acceleration and then use x-t relation: (delta x) = (initial v)(delta t) + (.5a(delta t)^2) to find displacement...