grade 11 physics

PLEASE HELP - NOBODY ANSWERED MY QUESTION YESTERDAY.

sue drives her 1452 kg car along a straight road at a constant velocity of 30 m/s [E]. Ger brakes suddently give out. she puts the car in neutral and lets it coast for 25 seconds. The air drag decelerates the car to a velocityof 25 m/s [E]. assume a frictionless surface.

PART C IS WHERE IM CONFUSED, BUT YOU'LL NEED TO KNOW MY ANSWERS TO THE OTHER ONES FIRST.

b)determine average acceleration while car's decelerating

m=1452 kh
v1=30 m/s [E]
v2=25 m/s [E]
t=25s

d=1/2(v1+v2)(t)
=1/2(55)(25)
=688 m

a= v2squared-v1squared/2(d)
=25squared-30squared/2(688)
= -0.20 m/ssquared

b)determine average force of air against car

F=(m)(a)
=(1452)(-.2)
=-290.4N

c)after coasting for 25s, she pulls her e-brake to slow to the car to a stop. If it takes 3s to stop the car, what is the force applied by the e-brake? Assume that the force exerted by the air remains constant and is equal to the forst determined in part (b).

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  1. (a) Your acceleration could have been computed more quickly as
    a = [v(2) - v(1)] = -5.0 / 25 = -0.2 m/s^2

    Note that is it negative.

    (b) Force = ma = 1452*(-0.2) = -290 N
    (the minus sign means the force is backwards).

    (c) To decelerate a car in 3 s from a speed of 25 m/s, the acceleration must be -25/3 = -8.33 m/s^2
    -0.2 m/s^2 of that deceleration was due to the aerodynamic force, so the remaining -8.13 m/s^2 must be due to emergency brake friction.

    The force applied by the e-brake is
    F(eb) = m*a(eb) = -8.13*1452 = -12,100 N

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  2. Superwoman is hovering above the ground when a person free-falling goes by her at a terminal velocity of 140km/h. Unfortunately, the parachute does not open. Fortunately, superwoman is around. If it takes her 1.9s to realize the person is in distress, what must her acceleration be if she is to catch the parachutist just before she hits the ground 1000m below?

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