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).
To find the force applied by the e-brake, we can use Newton's second law of motion, which states that force (F) is equal to mass (m) times acceleration (a).
In this case, the car's mass (m) is given as 1452 kg. The acceleration (a) can be determined using the formula a = (v2 - v1) / t, where v2 is the final velocity (0 m/s) and v1 is the initial velocity (25 m/s). The time (t) taken to stop the car is given as 3 seconds.
a = (v2 - v1) / t
= (0 - 25) / 3
= -25 / 3
= -8.33 m/s²
Now we can calculate the force (F) applied by the e-brake using the formula F = m * a.
F = (mass)(acceleration)
= (1452 kg)(-8.33 m/s²)
= -12100.36 N
Therefore, the force applied by the e-brake is approximately -12100.36 N (negative because it acts in the opposite direction of motion).