What is the force applied by the e-brake to stop the car in 3 seconds?

In summary: Then solve for the new force.In summary, Sue drives her 1452 kg car along a straight road at a constant velocity of 30 m/s [E]. When her brakes give out, she puts the car in neutral and lets it coast for 25 seconds, decelerating to a velocity of 25 m/s [E] due to air drag on a frictionless surface. The average acceleration while the car is decelerating is -0.20 m/ssquared and the average force of air against the car is -290.4N. When Sue pulls the e-brake to stop the car after coasting for 25 seconds, it takes 3 seconds and the force applied by the e-brake is equal
  • #1
Shaley
12
0
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 let's 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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  • #2
In part c there are two forces providing negative acceleration. You already know one from part b. To do part c first find the total acceleration. Call it a. Then say that the known force plus the force you are looking for are mass times the acceleration that you just found.
 
  • #3


To determine the force applied by the e-brake, we can use Newton's second law: F = ma. Since the car is already decelerating at -0.2 m/ssquared due to air resistance, the e-brake needs to provide an additional force to bring the car to a complete stop in 3 seconds. Therefore, the force applied by the e-brake can be calculated as:

F = m x a
= 1452 kg x (-0.2 m/ssquared)
= -290.4 N

This means that the e-brake needs to apply a force of -290.4 N in the opposite direction of the car's motion to bring it to a stop in 3 seconds. Keep in mind that this is an ideal scenario and in reality, the force applied by the e-brake may need to be greater due to other factors such as friction and the weight of the car. It is important to always practice safe driving techniques and maintain your vehicle's braking system to avoid any potential accidents.
 

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