Calculus AB: Applications of the Derivative

v(t) = s'(t) = 4t - 6
v(1) = - 2 feet per second
v(4) = 10 feet per second
Thus, the particle changes direction during this time interval.

It would not be correct to simply take s(4) - s(1) (the net change in position) in this case because the object spends part of the time moving forward, and part of the time moving backwards. When the object doubles back on itself, that overlapping distance is not captured by the net change in position. To solve this problem, we must add up the distances that the object travels when it is moving in one direction. To do this, we must find the point at which it changes direction. This occurs when the instantaneous velocity is equal to zero.
v(t) = 4t - 6
v(t) = 0 at x =
So, the object moves backward on (1,) and forwards on (, 4) . The distance traveled on the first interval is

s() - s(1) = -3 - (- 6) = 3 feet .

The distance traveled on the second interval is

s(4)-s() = 10-(-3) = 13 feet .

The total distance traveled is the sum of these two, which is 16 feet.

Here, acceleration is in the negative direction, since it is toward the ground, which is designated with a coordinate of zero. So, a = - 10 meters per second per second, and v(t) = v ₀ + a(t) . But the initial velocity is zero, so v(t) = - 10t . v(2) = - 20 meters per second.

s ₀ = 100 , so
s(t) = 100 + (- 10)t ²
s(t) = 100 - 5t ²
s(2) = 100 - 20
s(2) = 80 meters from the ground