can any one help me with that 10 points
A car is traveling at a constant velocity of 10 m/s for 10 seconds. The driver notices a lady crossing the street 20 meters away, and immediately applies the brakes (accelerates at -2 m/s). Does he hit the lady?
A car is traveling at a constant velocity of 10 m/s for 10 seconds. The driver notices a lady crossing the street 20 meters away, and immediately applies the brakes (accelerates at -2 m/s). Does he hit the lady?
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If he is moving at 10 meters per second and his brakes provide a deleration rate
of 2 m per sec per sec (which is what I believe you mean by -2 m/s; it's -2 m/sec²)
it will take him 5 seconds to come to a complete stop. Let's see how far he will
travel in 5 seconds while decelerating from 10 m/sec to a stop.
The distance d = ½at² where a is the acceration rate (here -2 m/sec²) and we
found the time t to be 5 seconds. Then the distance d in meters to stop is:
d = ½(2)(25) = 25 meters.
If the lady is only 20 meters away he will hit her unless she gets out of the way!
of 2 m per sec per sec (which is what I believe you mean by -2 m/s; it's -2 m/sec²)
it will take him 5 seconds to come to a complete stop. Let's see how far he will
travel in 5 seconds while decelerating from 10 m/sec to a stop.
The distance d = ½at² where a is the acceration rate (here -2 m/sec²) and we
found the time t to be 5 seconds. Then the distance d in meters to stop is:
d = ½(2)(25) = 25 meters.
If the lady is only 20 meters away he will hit her unless she gets out of the way!
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vo = 10 m/s
s = 20 m
a = 2 m/s^2
s = ½ at^2 so t = Sqrt(2s/a) = Sqrt(40/2) = 4.472 sec
After braking for 4.472 sec over distance of 20 m, the speed is:
v = vo - at = 10 - (2 x 4.472) = 1.146 m/s - final speed
Yes, he hits the lady.
s = 20 m
a = 2 m/s^2
s = ½ at^2 so t = Sqrt(2s/a) = Sqrt(40/2) = 4.472 sec
After braking for 4.472 sec over distance of 20 m, the speed is:
v = vo - at = 10 - (2 x 4.472) = 1.146 m/s - final speed
Yes, he hits the lady.