a driver traveling at a speed of 21m/s tops a hill and spots a deer standing in the middle of the road 90 meters away. he hits the breaks and panic stops the car in 8 seconds. how far did the car travel before stopping
+8581 Assuming that his acceleration was linear (ie. that he slowed down at a perfectly steady rate), then you can say his velocity at any point in those eight seconds was:
21 - t(21 / 8) m/s
To get the distance covered in that time, you can simply take the area under that line between 0 seconds and eight seconds. Because it's linear, this is very easy. You're dealing with a right triangle that is 8 seconds wide and 21 meters per second tall. With a triangle, the area is equal to half it's width times it's height, so the equation is:
d = 21m/s * 8s / 2
∴d = 168m / 2
∴d = 84m.
So he traveled 84 meters in those 8 seconds, and probably scared the bejeebers out of that poor deer.
Average speed = 21/2 m/s , Time =8 s, Distance = 8 x 21/2 = 84m
That poor dear must of had a heart attack, haha.
+8581 Assuming that his acceleration was linear (ie. that he slowed down at a perfectly steady rate), then you can say his velocity at any point in those eight seconds was:
21 - t(21 / 8) m/s
To get the distance covered in that time, you can simply take the area under that line between 0 seconds and eight seconds. Because it's linear, this is very easy. You're dealing with a right triangle that is 8 seconds wide and 21 meters per second tall. With a triangle, the area is equal to half it's width times it's height, so the equation is:
d = 21m/s * 8s / 2
∴d = 168m / 2
∴d = 84m.
So he traveled 84 meters in those 8 seconds, and probably scared the bejeebers out of that poor deer.
Average speed = 21/2 m/s , Time =8 s, Distance = 8 x 21/2 = 84m
That poor dear must of had a heart attack, haha.
