The height of an object thrown vertically upward is given by the function s=−4.9t2+20t+2 , where s is in meters and t is in seconds. What is the object’s velocity at t = 3 seconds? (1 point) Responses −5.3 m/s − 5.3 m/s 5.3 m/s 5.3 m/s −9.4 m/s − 9.4 m/s 9.4 m/s 9.4 m/s 17.9 m/s
To find the velocity at a given time, we need to find the derivative of the height function with respect to time.
Taking the derivative of s = -4.9t^2 + 20t + 2, we get:
s' = -9.8t + 20
Now we can substitute t = 3 into the expression for s' to find the velocity at t = 3 seconds:
s'(3) = -9.8(3) + 20 = -29.4 + 20 = -9.4 m/s
Therefore, the object's velocity at t = 3 seconds is -9.4 m/s.
To find the velocity of the object at t = 3 seconds, we need to find the derivative of the equation s = -4.9t^2 + 20t + 2 with respect to t.
The derivative of s with respect to t is given by ds/dt.
ds/dt = -9.8t + 20
Now, substitute t = 3 into the derivative equation to find the velocity:
ds/dt = -9.8(3) + 20
= -29.4 + 20
= -9.4 m/s (rounded to one decimal place)
Therefore, the object’s velocity at t = 3 seconds is -9.4 m/s. So, the correct answer is: −9.4 m/s.
To find the velocity at a specific time, you need to find the derivative of the height function with respect to time. In this case, the height function is given by:
s = -4.9t² + 20t + 2
To find the derivative, differentiate each term of the function with respect to time:
ds/dt = d/dt(-4.9t²) + d/dt(20t) + d/dt(2)
= -4.9 * d/dt(t²) + 20 * d/dt(t) + 0
= -4.9 * 2t + 20 * 1
= -9.8t + 20
Now, you have the derivative of the height function, which represents the velocity of the object at any given time.
To find the velocity at t = 3 seconds, substitute t = 3 into the derivative function:
v = -9.8(3) + 20
= -29.4 + 20
= -9.4 m/s
Therefore, the object's velocity at t = 3 seconds is -9.4 m/s. Thus, the correct answer is: -9.4 m/s.