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. (1 point) Responses −27 m/s − 27 m / s −29 m/s − 29 m / s −31 m/s − 31 m / s −33 m/s − 33 m / s −35 m/s
To find the object's velocity at t=6 seconds, we need to find the derivative of the height equation with respect to time:
y(t) = 2250 - 50t - 250e^(-0.2t)
Taking the derivative of y(t) with respect to t, we get:
dy/dt = -50 + 50(0.2)e^(-0.2t) + 50e^(-0.2t)
Now we can substitute t=6 into the derivative equation to find the velocity at t=6:
dy/dt = -50 + 50(0.2)e^(-0.2(6)) + 50e^(-0.2(6))
Simplifying,
dy/dt = -50 + 50(0.2)e^(-1.2) + 50e^(-1.2)
Now we can calculate dy/dt:
dy/dt = -50 + 50(0.2)(0.30119421191) + 50(0.30119421191)
dy/dt ≈ -50 + 3.0119421191 + 15.059710595 ≈ -31.928057881
Rounding the answer to the nearest whole number, the object's velocity at t=6 seconds is -32 m/s.
Which of the following best represents the magnitude of the object’s acceleration at t = 2.5 s? Round your answer to the nearest tenth.
To find the object's acceleration at t = 2.5 s, we need to find the second derivative of the height equation:
y(t) = 2250 - 50t - 250e^(-0.2t)
Taking the derivative of y(t) with respect to t, we obtain:
dy/dt = -50 + 50(0.2)e^(-0.2t) + 50e^(-0.2t)
Next, we take the derivative of dy/dt with respect to t to find the acceleration:
d^2y/dt^2 = 50(0.2)(-0.2)e^(-0.2t) + 50(-0.2)e^(-0.2t)
Simplifying,
d^2y/dt^2 = -20(0.2)e^(-0.2t) - 10e^(-0.2t)
Now we can substitute t = 2.5 into the acceleration equation:
d^2y/dt^2 = -20(0.2)e^(-0.2(2.5)) - 10e^(-0.2(2.5))
Simplifying,
d^2y/dt^2 ≈ -4e^(-0.5) - 10e^(-0.5)
Using a calculator, we find that e^(-0.5) ≈ 0.60653066.
d^2y/dt^2 ≈ -4(0.60653066) - 10(0.60653066) ≈ -2.42612264 - 6.0653066 ≈ -8.49142924
Rounding the answer to the nearest tenth, the magnitude of the object's acceleration at t = 2.5 s is approximately 8.5 m/s².