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 0.
a)
when is v = 0?
(t-5)(t-2) = 0
t = 5 or 2
b)
a = 2 t - 7
put in t = 5 and t = 2
c)
x = (1/3) t^3 - (7/2) t^2 + 10 t + c
put in given point to find c
d)
when is x a minimum?
When v = 0
so put in t = 5 and t = 2 and find the lower x value
To find the times when the object is at rest, we need to set the velocity equation equal to zero and solve for t.
(a) v(t) = t^2 - 7t + 10 = 0
To solve this equation, we can either factor it or use the quadratic formula. Let's factor it:
(t - 2)(t - 5) = 0
So, t - 2 = 0 or t - 5 = 0.
If t - 2 = 0, then t = 2. If t - 5 = 0, then t = 5.
Therefore, the object is at rest at t = 2 and t = 5.
(b) To find the acceleration at these times, we need to take the derivative of the velocity equation with respect to t.
a(t) = v'(t) = 2t - 7 m/s^2
Substituting t = 2 and t = 5 into the acceleration equation, we get:
a(2) = 2(2) - 7 = -3 m/s^2
a(5) = 2(5) - 7 = 3 m/s^2
Now, let's interpret the acceleration. A negative acceleration (-3 m/s^2) means that the object is slowing down since its speed is decreasing. A positive acceleration (3 m/s^2) means that the object is speeding up since its speed is increasing. The direction of the object's motion cannot be determined solely from the given information.
(c) To find the position of the object at time t, we need to integrate the velocity equation with respect to t.
x(t) = ∫(t^2 - 7t + 10) dt
= (1/3)t^3 - (7/2)t^2 + 10t + C
Since the initial position is given as x = -1 at t = 1, we can substitute these values into the equation:
-1 = (1/3)(1)^3 - (7/2)(1)^2 + 10(1) + C
Simplifying this equation, we get:
-1 = 1/3 - 7/2 + 10 + C
C = -1 - 1/3 + 7/2 - 10
C = 49/6 - 1/3 - 60/6
C = -1/6
Therefore, the position equation becomes:
x(t) = (1/3)t^3 - (7/2)t^2 + 10t - 1/6
To find the position of the object at any given time t, substitute the value of t into the position equation.
(d) To find the leftmost position of the object for all t ≥ 0, we need to find the minimum value of the position equation.
To find the minimum, we can take the derivative of the position equation with respect to t and set it equal to zero.
x'(t) = (1/3)(3t^2) - (7/2)(2t) + 10
= t^2 - 7t + 10 = 0
We have already determined that this quadratic equation has roots at t = 2 and t = 5. Since the coefficient of t^2 is positive, the parabola opens upwards, so the minimum value occurs between these two roots.
To find the leftmost position, we substitute t = 0, t = 2, and t = 5 into the position equation:
x(0) = (1/3)(0)^3 - (7/2)(0)^2 + 10(0) - 1/6 = -1/6
x(2) = (1/3)(2)^3 - (7/2)(2)^2 + 10(2) - 1/6 = 5/3
x(5) = (1/3)(5)^3 - (7/2)(5)^2 + 10(5) - 1/6 = 30/3 = 10
The leftmost position occurs at t = 2, which corresponds to x = 5/3.