# physics

Calculate the force needed to bring a 1070–kg car to rest from a speed of 89.0 km/h in a distance of 110 m (a fairly typical distance for a non-panic stop).

Suppose instead the car hits a concrete abutment at full speed and is brought to a stop in 2.34 m. Calculate the force exerted on the car and compare it with the force found in part (a).

Force = -1.40×105 N
Ratio = 47.0

what are the steps to get these answers?

1. 0
2. 5
1. force = mass * acceleration

now the kinematics

first units
89 km/h * 1000 meters/km * 1 h/3600 seconds
= 24.7 meters/second

average speed during stop = 24.7/2 = 12.36
so how long did it take?
t = 110 meters/12.36 m/s
= 8.9 seconds to stop

acceleration assumed constant so it is
a = -24.7 m/s / 8.9 s
a = -2.78 m/s^2 (about 1/3 acceleration of gravity)

F = -1070*2.78 = -2975 N
so we agree

Now do all that for 2.34 meters (OUCH!!)
t = 2.34 meters/12.36 m/s
= .1893 seconds
a = -24.7/.1893 = - 130 m/s^2
( about 13 times gravity !!! )
F = -1070 * 130 = -139,599 N

139599/2975 = 46.9

1. 0
posted by Damon

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