Question

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 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?

Answer 1

Answer:

2.73414 seconds

467622.66798 J

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

g = Acceleration due to gravity = 9.81 m/s² = a

$v=60\times \dfrac{1609.34}{3600}=26.822\ m/s$

$mgh=\dfrac{1}{2}mv^2\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow h=\dfrac{26.822^2}{2\times 9.81}\\\Rightarrow h=36.66766\ m$

$s=ut+\frac{1}{2}at^2\\\Rightarrow 36.66766=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{36.66766\times 2}{9.81}}\\\Rightarrow t=2.73414\ s$

or

$v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{26.822-0}{9.81}\\\Rightarrow t=2.73414\ s$

The time taken is 2.73414 seconds

The potential energy is given by

$U=mgh\\\Rightarrow U=1300\times 9.81\times 36.66766\\\Rightarrow U=467622.66798\ J$

The change in potential energy is 467622.66798 J

Answer 2

A) The time the magazine car take to go from zero to 60 mph is; t = 2.737 s

B) The magnitude of the change in gravitational potential energy during the given time is; ΔGPE = 467,685.4 J

We are given;

Speed; v = 60 mph = 26.8224 m/s

Mass of car; m = 1300 kg

Time; t = 6 s

From principle of conservation of energy, we can say that;

mgh = ½mv²

Thus;

h = v²/2g

Plugging in the relevant values;

h = 26.8224²/(2 × 9.8)

h = 36.71 m

We can find how long would it take to go from zero to 60 mph using Newton's first equation of motion. Thus;

t = (v - u)/g

t = (26.8224 - 0)/9.8

t = 2.737 s

Now, formula for magnitude of change in gravitational potential energy is;

ΔGPE = mgh

ΔGPE = 1300 × 9.8 × 36.71

ΔGPE = 467,685.4 J

Read more at; https://brainly.com/question/25708521