Answer :
Final answer:
To determine the drag force needed to slow an 892-kg car down to 50.0 mi/h over a distance of 186m, first convert the speed to m/s and find the change in velocity. Then, use a kinematic equation to find the acceleration, and finally multiply this acceleration by the car's mass to find the force. The force is approximately -1193.28 N.
Explanation:
The question pertains to the concept of Force in physics, specifically in context of a phenomenon called 'drag'.
To find the necessary drag force, we first need to calculate the change in velocity and then use the formula for Force which is mass x acceleration.
In this case, the car needs to go from its current speed to 50.0 mi/h, which is approximately 22.35 m/s.
If we assume the car is initially at rest, then the change in velocity (∆v) is 22.35 m/s.
Next, we can calculate acceleration, remembering that acceleration is change in velocity over time (or in this case, over a certain distance).
Substituting the given values into the kinematic equation v^2 = u^2 + 2as (where 'v' is the final velocity, 'u' is the initial velocity, 'a' is the acceleration, and 's' is distance), we find the acceleration to be around -1.34 m/s².
Finally, to find the drag force, we multiply the mass of the car by the acceleration (F = ma).
So, the force exerted by the drag chute to slow down the car over a distance of 186m should be around -1193.28 N.
The force needed to slow down the car can be determined using the principles of physics. To find the force exerted by the drag chute, we can utilize the equation of motion.i2 +2a⋅d
Where:
final velocity
=50mi/hv f
=final velocity=50mi/h
initial velocity
=initial velocity of the car
=v i
=initial velocity=initial velocity of the car=? (It's not provided)
=deceleration
a=deceleration (This will be calculated)
d=distance=186m
Firstly, we need to convert the final velocity from miles per hour to meters per second since the distance is given in meters.
Final velocity=50.0mi/h
Initial velocity=?
We'll start by converting the final velocity from miles per hour to meters per second.
=50.0mi/h×( 3600s1609m )≈22.35m/s
The initial velocity of the car is not provided, and we'll assume the car starts from rest (=0v i =0).
Now, we rearrange the equation to solve for the deceleration (
a= 2dv f2−v i2
Since the car starts from rest ( =0), the equation becomes:
=(22.35 m/s)22
186m
a= 2dV f2
= 2×186m(22.35m/s)
a= 372m500.1225m 2/s 2
a≈1.344m/s 2
Next, to find the force (F) exerted by the drag chute, we can use Newton's second law of motion,
F=m⋅a, where
m is the mass of the car:
=892 kg×1.344m/s2
F=m⋅a=892kg×1.344m/s 2
≈1200.448 NF≈1200.448N
Therefore, the force that the drag chute must exert on the 892 kg car to slow it to 50.0 mi/h in a distance of 186 m is approximately 1200.45 Newtons.
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