The position of an object changes with time according to the expression x(t) = 5t62 = 3t - 1. What is the position of the object when its velocity is 11 m/s?
To find the position of the object when its velocity is 11 m/s, we first need to determine the instantaneous velocity at any given time. The velocity can be found by taking the derivative of the position function with respect to time, dx/dt.
Given the position function x(t) = 5t^2 + 3t - 1, we can differentiate it to find the velocity function:
v(t) = dx(t)/dt = d/dt (5t^2 + 3t - 1)
Differentiating each term separately:
v(t) = d/dt (5t^2) + d/dt (3t) + d/dt (-1)
Using the power rule of differentiation, we get:
v(t) = 10t + 3
Now that we have the velocity function v(t) = 10t + 3, we can set it equal to 11 m/s and solve for the corresponding time:
10t + 3 = 11
Subtracting 3 from both sides:
10t = 8
Dividing both sides by 10:
t = 0.8
Now that we know the time when the velocity is 11 m/s is t = 0.8, we can substitute this value back into the original position function to find the position:
x(t) = 5t^2 + 3t - 1
x(0.8) = 5(0.8)^2 + 3(0.8) - 1
Calculating the expression:
x(0.8) = 5(0.64) + 2.4 - 1
x(0.8) = 3.2 + 2.4 - 1
x(0.8) = 4.6
Therefore, the position of the object when its velocity is 11 m/s is 4.6 units.