An object is traveling along the x axis so that its speed is given by v(t)=t^2-7t+10 m/s.
(a) Find the times when the object is at rest.
(b) Find the acceleration at both of these times. Interpret the acceleration in
terms of the motion of the object, ie whether the object is increasing or
decreasing in speed and in which direction.
(c) If the object is at x=-1 m at t=1, find the position of the object at time t.
(d) Find the leftmost position of the object for all t > or equal to 0.
To find the times when the object is at rest, we need to find the values of t for which the velocity equals zero.
(a) Set the velocity equation v(t) = t^2 - 7t + 10 equal to zero:
t^2 - 7t + 10 = 0
We can solve this equation using factoring or the quadratic formula. Factoring gives us:
(t - 2)(t - 5) = 0
From this, we can see that the object is at rest when t = 2 or t = 5.
(b) To find the acceleration at these times, we need to take the derivative of the velocity function with respect to time.
The velocity function is v(t) = t^2 - 7t + 10.
Taking the derivative, we get:
a(t) = 2t - 7
Now substitute t = 2 and t = 5 into the acceleration equation:
a(2) = 2(2) - 7 = -3 m/s^2
a(5) = 2(5) - 7 = 3 m/s^2
The negative acceleration at t = 2 indicates that the object is decreasing in speed and changing direction. The positive acceleration at t = 5 indicates that the object is increasing in speed in the positive direction.
(c) To find the position of the object at any given time t, we need to integrate the velocity function with respect to time.
The integral of v(t) = t^2 - 7t + 10 is given by:
x(t) = ∫(t^2 - 7t + 10) dt
Evaluating the integral, we get:
x(t) = (1/3)t^3 - (7/2)t^2 + 10t + C
Given that the object is at x = -1 m when t = 1, we can substitute these values into the equation to solve for the constant C:
-1 = (1/3)(1)^3 - (7/2)(1)^2 + 10(1) + C
Simplifying the equation, we find:
C = -1 - 1/3 + 7/2 - 10
C = 1/6
Now we can rewrite the position equation with the value of C:
x(t) = (1/3)t^3 - (7/2)t^2 + 10t + 1/6
(d) To find the leftmost position of the object for all t ≥ 0, we need to determine the minimum value of the position function.
Taking the derivative of x(t) with respect to t and setting it equal to zero, we can find the critical points where the leftmost position occurs:
x'(t) = 3t^2 - 7t + 10 = 0
This equation does not have real roots, which means there are no critical points within the given time range. Therefore, the leftmost position is the initial position (when t = 0), which is given by:
x(0) = (1/3)(0)^3 - (7/2)(0)^2 + 10(0) + 1/6
x(0) = 1/6
Hence, the leftmost position of the object for all t ≥ 0 is at x = 1/6.