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?
Take the derivative of x(t) with respect to t to find a function v(t). Set that equal to 11, solve for t, then put that in to x(t) and compute x.
Do you mean x(t) = 5t^2 +3t -1?
What you wrote makes no sense.
If you meant what I wrote, the velocity at time t is
V = x'(t) = 10 t + 3
When x'(t) = 11,
11 = 10 t + 3
10 t = 8
t = 0.8
use x(t) to compute x at that time
To determine the position of the object when its velocity is 11 m/s, we need to first find the time(s) at which the velocity is 11 m/s. Then, we can plug that time(s) into the position equation to find the corresponding position(s) of the object.
Given the position expression x(t) = 5t^2 + 3t - 1, we can find the velocity expression by taking the derivative of x(t) with respect to time (t):
v(t) = d(x(t))/dt = d(5t^2 + 3t - 1)/dt.
Differentiating each term separately, we get:
v(t) = 10t + 3.
Now, we need to solve the equation 10t + 3 = 11 to find the time(s) at which the velocity is 11 m/s. Rearranging the equation, we get:
10t = 11 - 3,
10t = 8,
t = 8/10,
t = 0.8 seconds.
Now that we have the value of t, we can substitute it back into the position expression to find the position of the object at that time:
x(t) = 5t^2 + 3t - 1,
x(0.8) = 5(0.8)^2 + 3(0.8) - 1,
x(0.8) = 5(0.64) + 2.4 - 1,
x(0.8) = 3.2 + 2.4 - 1,
x(0.8) = 4.8 - 1,
x(0.8) = 3.8 meters.
Therefore, the position of the object when its velocity is 11 m/s is 3.8 meters.