In magazine car tests an important indicator of performance is the zero to 60 mph 0 to 96.6 km/h) acceleration time. A time below 6.0 s is considered to be extremely quick. If a car with mass 1300 kg were dropped vertically from a great height, how long would it take to go from zero to 60 mph and what would the magnitude of the change in gravitational potential energy be during this time? neglect air resistance)
A) 2.74 s and 17.4 kJ
B) 2.74 s and 468 kJ
C) 2.74 s and 468 mJ
D) 0.37 s and 468 mJ
E) 2.74 s and 17.4 mJ

Question

Quizplus · Accepted Answer

To find the time it takes for the car to reach 60 mph, we can use the formula for acceleration due to gravity:

v = at

where v is the final velocity (60 mph or 26.8 m/s), a is the acceleration due to gravity (9.8 m/s^2), and t is the time it takes to reach that velocity.

Rearranging the formula, we get:

t = v/a = 26.8 m/s / 9.8 m/s^2 = 2.74 s

So it takes 2.74 seconds for the car to go from zero to 60 mph.

To find the change in gravitational potential energy, we need to use the formula:

ΔPE = mgh

where m is the mass of the car (1300 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height the car falls from.

Assuming the car is dropped from rest from a great height, h is unknown, but it cancels out when we use conservation of energy:

ΔPE = ΔKE

where ΔKE is the change in kinetic energy, which we can find using:

KE = (1/2)mv^2

KE = (1/2) * 1300 kg * (26.8 m/s)^2 = 468,040 J or 468 kJ (to three significant figures)

Therefore, the magnitude of the change in gravitational potential energy is also 468 kJ, so the best answer choice is (B).

In Magazine Car Tests an Important Indicator of Performance Is