The driver of a car applies the brakes when he sees a tree that has fallen on the road, and t seconds later the distance, in meters, from the car to the tree is x=2t^2 -16t +35 (until the car comes to a stop, after which the equation is no longer valid.) How far does the car travel before it stops? Does the car hit the tree?
+37170 If we take dx/dt of the x= distance equation we will get the velocity = 4t-16 When the car comes to a stop, the velocity = 0 , so setting this equation = 0 will give us t in seconds when the car stops:
4t-16 = 0 yields car stop at t= 4 seconds NOW, going back to the DISTANCE BETWEEN equation with t= 4
x= 2t^2 - 16t +35 yields 32 -64 +35 = 3 meters between the car and the tree...so the car will NOT hit the tree, but stop 3 meters from it.
The distance the car travels is the difference between t= 0 and t= 4 for the distance between equation given
t=0 yields 35 t= 4 we found = 3 35-3 = 32 meters = distance car travelled
+37170 If we take dx/dt of the x= distance equation we will get the velocity = 4t-16 When the car comes to a stop, the velocity = 0 , so setting this equation = 0 will give us t in seconds when the car stops:
4t-16 = 0 yields car stop at t= 4 seconds NOW, going back to the DISTANCE BETWEEN equation with t= 4
x= 2t^2 - 16t +35 yields 32 -64 +35 = 3 meters between the car and the tree...so the car will NOT hit the tree, but stop 3 meters from it.
The distance the car travels is the difference between t= 0 and t= 4 for the distance between equation given
t=0 yields 35 t= 4 we found = 3 35-3 = 32 meters = distance car travelled
+37170 Another option is to graph (plot) the x =2t^2-16t+35 and you will se that the graph never reaches zero..(i.e. the car never reaches the tree)....the low point (nadir) is at t=4 with a value of 3 (as found in the previous solution)..
