Asked by Basharat
A baloon is 200 m off the ground and rising vertically at the rate of 15 m/sec an automobile passes beneath it travelling along a straight road at a constant rate of 45km/hour how fast is the distance between them changing one second later???
Answers
Answered by
Steve
after t seconds, the distance d is
d^2 = (200+15t)^2 + (12.5t)^2
d^2 = 381.25t^2 + 960t + 6400
So,
2d dd/dt = 762.5t + 960
At t=1, d=215.36
So, at t=1, dd/dt = 4.0 m/s
d^2 = (200+15t)^2 + (12.5t)^2
d^2 = 381.25t^2 + 960t + 6400
So,
2d dd/dt = 762.5t + 960
At t=1, d=215.36
So, at t=1, dd/dt = 4.0 m/s
Answered by
Steve
No, that can't be right. The balloon is rising at 15 m/s.
Aha.
d^2 = 381.25t^2 + 6000t + 40000
2d dd/dt = 762.5t + 6000
dd/dt = 15.7 m/s
Aha.
d^2 = 381.25t^2 + 6000t + 40000
2d dd/dt = 762.5t + 6000
dd/dt = 15.7 m/s
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