The height y of an object falling from rest from a height of 2,000 m is given by the equation y=2250−50t−250e−0.2t , where t is in seconds. What is the object’s velocity at t = 6 s? Round your answer to the nearest whole number.
To find the velocity of the object at t = 6 s, we need to find the derivative of the equation with respect to time.
Differentiating the equation y = 2250 - 50t - 250e^(-0.2t) with respect to t, we get:
dy/dt = -50 - 250(-0.2)e^(-0.2t)
At t = 6 s, the velocity is given by dy/dt at t = 6:
dy/dt = -50 - 250(-0.2)e^(-0.2*6)
dy/dt = -50 + 250(-0.2)e^(-1.2)
dy/dt ≈ -50 + 250(-0.2)(0.301194)
dy/dt ≈ -50 + 250(-0.060239)
dy/dt ≈ -50 - 15.05975
dy/dt ≈ -65.05975
Rounding to the nearest whole number, the object's velocity at t = 6 s is approximately -65 m/s.