SOLUTION: An auto A is moving at 30 fps and accelerating at 5 fps^2 to overtake an auto B
which is 384 ft ahead. Auto B is moving at 60 fps and decelerating at 3 fps^2
(a) How soon will A
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Question 1165401: An auto A is moving at 30 fps and accelerating at 5 fps^2 to overtake an auto B
which is 384 ft ahead. Auto B is moving at 60 fps and decelerating at 3 fps^2
(a) How soon will A pass B?
(b) What is the total distance traveled by car A?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
i believe the distance formula for an accelerating car is y = ax + (b/2)x^2 and the distance formula for a decelerating car is y = ax - (b/2)x^2
x is the number of seconds.
y is the distance traveled in x seconds.
a is the constant velocity
b is the constant constant acceleration.
applying this to your situation gets you.
y = 30 * x + 5/2 * x^2 for the accelerating car.
y = 60 * x - 3/2 * x^2 for the decelerating car.
both cars will have traveled the same distance when 30 * x + 5/2 * x^2 = 60 * x - 3/2 * x^2
solve for x to get x = 7.5 seconds.
both cars will have traveled the same distance in 7.5 seconds, after w hich the accelerating car will have passed the decelerating car.
at this point in time, the distanced traveled by each car will be 365.625 feet.
the equations for each car can be graphed.
the intersection of the lines on the graph is when both cars will have traveled the same distance.
the equation for the accelerating car is y = 30x + 5/2 * x^2.
the equation for the decelerating cvar is y = 60x - 3/2 * x^2.
the graph confirms the solution expressed above (365.625 feet in 7.5 seconds for both cars).
i also used a calculator to confirm the formulas were correctly applied.
for the accelerating car .....
for the decelerating car .....
both the graph and the calculator confirm the calculations are correct.
this assumes, of course, that i'm using the right formula.
i'm reasonably confident that i am.
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