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Wize AP Calculus (AB) Textbook > Applications of Integration

Position, Velocity, and Acceleration

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Position, Velocity, and Acceleration

Let s(t)s(t) be the position of an object at time tt where tt is in seconds and t≥0t \geq 0. Recall from derivatives that the velocity of the object at time tt is found as
v(t)=s′(t)=dsdt\boxed{\displaystyle v(t) = s'(t)=\frac{ds}{dt}}

and the acceleration of the object at time tt is found as
a(t)=v′(t)=dvdt=s′′(t)\boxed{\displaystyle a(t)=v'(t)=\frac{dv}{dt}=s''(t)}


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Velocity and Position from Integrals

From what we now know of antiderivatives, this means that we can find velocity and position of the object given only its acceleration, its initial velocity v0v_0, and its initial position s0s_0:
v(t)=v0+∫a(t) dt\boxed{v(t)=v_0+\int a(t)\ dt }

s(t)=s0+∫v(t) dt\boxed{s(t)=s_0+\int v(t) \ dt}

Wize Concept
v0v_0 and s0s_0 are the constants of integration (the +C+C from indefinite integrals).

  • Speed is the absolute value of velocity, ∣v(t)∣|v(t)|
  • If an object is dropped from rest, its only acceleration is due to gravity, g≈9.8 m/s2g\approx 9.8 \ \text{m/s}^2

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Example: Position and Velocity

Given that the velocity of a body, starting from the origin, and moving along the x-axis, is v(t)=9.8t−5\displaystyle v(t)=9.8t-5, find the position of the moving body at t=3t=3.

Since the velocity is the derivative of position, position will be the integral of the velocity:

v(t)=dsdt=9.8t−5\displaystyle v\left(t\right)=\frac{ds}{dt}=9.8t-5

Integrate to find position:

s(t)=∫(9.8t−5) dt\displaystyle s(t)=\int{(9.8t-5)}\ dt

=9.82 t2−5t+C\displaystyle =\dfrac{9.8}{2}\ t^2-5t+C

Find the constant using the fact that the object starts at the origin, that is, position is zero when time is zero: s(0)=0s(0)=0. Put this in the equation to get:

0=9.82 (0)2−(0)t+C\displaystyle 0=\dfrac{9.8}{2} \ (0)^2-(0)t+C

⇒C=0\Rightarrow C=0

Therefore the position function is:

s(t)=9.82 t2−5t\displaystyle s(t)=\dfrac{9.8}{2}\ t^2-5t

At t=3t=3 the position is:

s(3)=(9.8)(9)2−5(3)s(3)=\dfrac{(9.8)(9)}{2}-5(3)

=29.1=\boxed{29.1}

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Example: Position, Velocity, and Acceleration

Suppose a stone is dropped from a 100-m tall cliff. Ignoring air resistance, with what speed will it hit the ground?

If we take up to be positive, then the ground is where s(t)=0s(t)=0, and s0=100s_0=100. Since the stone is dropped from rest, we also have that v0=0v_0=0. The only acceleration is due to gravity, and since the object is traveling downward, we take a=−9.8a=-9.8 (in meters per second squared).
Given all this, we wish to find v(t)v(t) at the moment that s(t)=0s(t)=0.

We first find v(t)=v0+∫a(t) dtv(t)=v_0 + \int a(t)\ dt:
v(t)=v0+∫a(t) dt=0+∫−9.8 dt=−9.8tv(t)=v_0 + \int a(t)\ dt =0 + \int -9.8\ dt = -9.8t
If we know the time at which the stone hits the ground, then we can find v at that time; but we don't. Our only information about hitting the ground is that it occurs when s(t)=0s(t)=0, so we'll find s(t)s(t):
s(t)=s0+∫v(t) dt=100+∫−9.8t dt=100+12(−9.8t2)=−4.9t2+100s(t)=s_0 + \int v(t)\ dt = 100 + \int-9.8t\ dt = 100+\dfrac{1}{2}(-9.8t^2)=-4.9t^2+100
We want to know the time at which s(t)=0s(t)=0:
s(t)=0→−4.9t2+100=04.9t2=100t2=100/4.9t=±100/4.9t≈±4.518\begin{array}{rl} s(t)=0 \rightarrow -4.9t^2+100 &= 0\\ 4.9t^2&=100 \\ t^2&=100/4.9\\ t&=\pm\sqrt{100/4.9}\\ t &\approx \pm 4.518 \end{array}
Since we're ignoring "negative time", we take t≈4.518t\approx 4.518, and find v(4.518)=−9.8(4.518)=−44.27v(4.518)=-9.8(4.518)=-44.27.
That is, the stone's speed at the instant it hit the ground was approximately 44.27 meters per second.
The position of an object is given by s(t)s(t). If its acceleration is given by a(t)=6t−2 m/s2a(t)=6t-2 \ \text{m/s}^2 and if it starts from s(0)=1 ms(0)=1 \text{ m} with a velocity v(0)=2 m/sv(0)=2 \ \text{m/s}, where will the object be at t=2t=2?