Calculus word problem

Question

An object is moving along a line, with time in seconds, and distance in feet. The acceleration of the object at timet is a(t) = -32 feet per second per second. The velocityof the object at t = 0 seconds is v(0) = 80 feet per second. The position of the object at time t = 0 seconds is s(0) = 10 feet. Find the velocity function v(t), and the position function s(t).

Answer 1

The velocity equation is the integral of the acceleration equation.

So v(t) = ∫-32 dt

This is an indefinite integral so when we integrate, we have to add "+C": v(t) = -32t + C

This allows us to use the information that when t = 0, v(0) = 80 ft/sec: v(0) = -32 (0) + C = 80 or 0 + C = 80 or C = 80.

So now we have our velocity equation: v(t) = -32t + 80 ft/sec.

The position equation is the integral of the velocity equation.

So s(t) = ∫(-32t + 80) dt

Again, this is an indefinite integral, so we integrate to get: s(t) = -32 (1/2) t² + 80t + C or s(t) = -16t² + 80t + C

Again, we use the information given in the problem that when t = 0, s(0) = 10 ft

So s(0) = 16 (0)² + 80 (0) + C = 10 or C = 10.

So now we have our position equation: s(t) = -16t² +80t + 10

You just have to memorize the relationship between position, velocity and acceleration.

position (derivative) -> velocity (derivative) -> acceleration

acceleration (integral) -> velocity (integral) -> position

Answer 2

The important things to know about this question is that acceleration is the change in velocity over the change in time (i.e. a = dv/dt), and velocity is the change in position of the change in time (i.e. v = ds/dt). Hence

1. find velocity
2. a = dv/dt
3. dv = adt
4. integrate: ∫dv = ∫adt = ∫-32dt
5. v = -32t + 80 = 80 - 32t ft/s. The 80 comes from the initial velocity.
6. find position
7. v = ds/dt
8. ds = vdt
9. integrate: ∫ds = ∫vdt = ∫(80 - 32t)dt
10. s = 80t - 32t²/2 + 10 ft = 10 + 80t - 32t²/2 ft. the 10 comes from the initial position.
11. Answers:
12. v = 80 - 32t ft/s
13. s = 10 + 80t - 32t²/2 ft