A 975-kg jeep slows to rest from a speed of 925 km/h in a distance of 125 m. Consider that the jeep is initially traveling in the positive direction. If the brakes are the only thing making the jeep come to a stop, calculate the force (in newtons, in a component along the direction of motion of the car) that the brakes apply on the jeep. Suppose instead of braking that the jeep hits a concrete abutment at full speed and is brought to a stop in 2.00 m. Calculate the force, in newtons, exerted on the jeep in this case. What is the ratio of the force on the jeep from the concrete to the braking force?

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boffeemadrid boffeemadrid · Answer 1 · 2020-01-30T10:50:15+01:00

Answer:

-257478 N

-16092483.225 N

62.5

Explanation:

u = Initial velocity = 925 km/h

v = Final velocity = 0

s = Displacement

a = Acceleration

[tex]v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-(\dfrac{925}{3.6})^2}{2\times 125}\\\Rightarrow a=-264.08\ m/s^2[/tex]

Force is given by

[tex]F=ma\\\Rightarrow F=975\times -264.08\\\Rightarrow F=-257478\ N[/tex]

The force applied is -257478 N

[tex]v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-(\dfrac{925}{3.6})^2}{2\times 2}\\\Rightarrow a=-16505.111\ m/s^2[/tex]

[tex]F=ma\\\Rightarrow F=975\times -16505.111\\\Rightarrow F=-16092483.225\ N[/tex]

The force applied is -16092483.225 N

The ratio is

[tex]\dfrac{-16092483.225}{-257478}=62.5[/tex]

The ratio is 62.5