Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

1- What is the minimum stopping distance for a 1300 k g car traveling 98 k m / h

ID: 2141120 • Letter: 1

Question

1- What is the minimum stopping distance for a 1300kg  car traveling 98km/h  if ?s = 0.55? Express your answer to two significant figures and include the appropriate units.

2- What would it be if the car were on the Moon (the acceleration of gravity on the Moon is about g/6) but all else stayed the same?          Express your answer to two significant figures and include the appropriate units.
1- What is the minimum stopping distance for a 1300kg  car traveling 98km/h  if ?s = 0.55? Express your answer to two significant figures and include the appropriate units.

2- What would it be if the car were on the Moon (the acceleration of gravity on the Moon is about g/6) but all else stayed the same?          Express your answer to two significant figures and include the appropriate units.
Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

since vf=0 for stopping, you have

v0^2/2a = d

you find the accel from newton's second law

F= ma; but here, the force comes from friction which is u mg where u is the coefficient of friction
therefore umg=ma or a = ug

use this expression for a and get:

d=v0^2/2ug notice the mass does not appear here, so the mass of the car is not relevant

d=v0^2/2ug

v0=98km/hr = 27.23m/s

d=27.23^2/2(0.55)(9.8) = 68.78 m

on the moon, g is 1/6 its earth's value, so the value of g in the denom goes down by a factor of six, so the distance increases by a factor of 6, and so d= 6x68.78m=412.68m