A car driving from Tucson to Nogales is distracted by some physics students laun

Question

A car driving from Tucson to Nogales is distracted by some physics students launching model rockets in the desert, veers off the road, and crashes into a cactus. The driver is luckily wearing his seatbelt, which causes him to decelerate from 110 km/hour to a dead stop in 50 milliseconds.

What is the average acceleration felt by the driver over this period?

How many times greater is this than the acceleration due to Earth’s gravity?

A driver not wearing a seatbelt would fly forward and strike the steering wheel, decelerating much more suddenly –- over, say, 6 ms. What is the acceleration (in "g’s", or multiples of Earth’s gravity) of the driver now?

Please show all work

Explanation / Answer

the car decelerates in 100 kmph in 50 ms

1 kmph = 5/18 m/s

100 kmph = 100*5/18 = 27.78 m/s

for calculating the decelration, we have to use the equation of linear motion

v = u + at

where v = final velcoity , u =initial velcoity , a= acceleration , t = time in s

0 = 27.78 + a*(50*10^-3)

a = 555.56 m/s2

average acceleration felt by the driver over this period = 555.56 m/s2

b) comparing it with the earth's acceleration due to gravity

= 555.56/9.8 = 56.7 times

c) if the driver decelerates in 6 ms

v = u +at

0 = 27.78 + a *(6*10^-3)

a = 4630 m/s2

a = 4630/9.8 = 472.44g m/s2

accelerartion in terms of g = 472.44g m/s2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.