An object is moving in a straight line according to s(t) = -t^2+4t-3 where s(t) = distance the object goes in miles at. t=times in hours
a. How far did the object go in 2 hours?
b. How fast was the object in 3 hours?
(a) s(2)-s(0) = (-4+8-3)-(-3) = 4 mi
(b) v(t) = ds/dt = -2t+4
v(3) = -2*3+4 = -2 mi/hr
From problem 1a. ^
a. What was the acceleration at 3 hrs?
b. When was the object farthest away?
a(t) = -2, so constant
max s is when v=0
To find the distance an object went in a given time or the velocity of the object at a specific time, we can use the given equation s(t) = -t^2 + 4t - 3.
a. To find how far the object went in 2 hours, we substitute t = 2 into the equation s(t) = -t^2 + 4t - 3:
s(2) = -(2)^2 + 4(2) - 3
s(2) = -4 + 8 - 3
s(2) = 1
Therefore, the object went 1 mile in 2 hours.
b. To find the velocity (speed) of the object at 3 hours, we need to find the derivative of the function s(t). The derivative of s(t) will give us the velocity of the object at any given time.
Taking the derivative of s(t) = -t^2 + 4t - 3:
s'(t) = -2t + 4
Now we substitute t = 3 into s'(t) to find the velocity at 3 hours:
s'(3) = -2(3) + 4
s'(3) = -6 + 4
s'(3) = -2
Therefore, the object had a velocity (speed) of -2 miles per hour at 3 hours. Note that the negative sign indicates that the object is moving in the opposite direction from the positive direction of the line.