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Wize University Calculus 1 Textbook > Applications of Integration for Physical Science

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?



Extra Practice
Applications of Integration: Position from Acceleration
The position of an object is given by x(t)x(t). If its acceleration is given by a(t)=6t−2a(t)=6t-2 m/s2 and if it starts from x(0)=1x(0)=1 m with a velocity v(0)=2v(0)=2m/s, where will the object be at t=2t=2?
Applications of Integration: Position from Acceleration
The position of an object is given by x(t)x(t). If its acceleration is given by a(t)=6t−2a(t)=6t-2 m/s2 and if it starts from x(0)=1x(0)=1 m with a velocity v(0)=2v(0)=2m/s, where will the object be at t=2t=2?
A clown is shot directly up out of a cannon with a velocity of 50 m/s50\ m/s. If gravity causes a −10 m/s2-10 \ m/s^2 constant acceleration, how high is the clown 22 seconds later?
Applications of Integration: Position, Velocity and Acceleration
Suppose a car hits the brakes at full power and comes to a complete stop after 50 meters. Assuming that the deceleration caused by the brakes is a constant −10 m/sec2-10 \text{ m/sec}^2, find the speed of the car (in meters per second, to two decimal places) when it began braking.
Applications of Integrals: Position, Velocity and Acceleration
Suppose a coin is dropped off the side of a 50 meter tall cliff. Find the speed of the coin when it hits the ground (to two decimal places) in meters per second.
Applications of Integration: Position, Velocity and Acceleration
If an object is dropped off a 100-meter cliff, find its average velocity (to two decimal places) over the time it is dropped to the time it hits the ground.
Applications of Integration: Position from Acceleration
The position of an object is given by x(t)x(t). If its acceleration is given by a(t)=6t−2a(t)=6t-2 m/s2 and if it starts from x(0)=1x(0)=1 m with a velocity v(0)=2v(0)=2m/s, where will the object be at t=2t=2?
Kinematics
The following graph represents the velocity of a particle in m/s, find the total displacement in meters of the particle from t=0 to t=5.
Related Rates
A clown is shot directly up out of a cannon with a velocity of
50 m/s50\ m/s. If gravity causes a −10 m/s2-10 \ m/s^2 constant acceleration, how high is the clown 22 seconds later?
Applications of Integration: Position, Velocity and Acceleration
A sky diver with a broken parachute falls at a constant acceleration
of −10 m/s2-10 \ m/s^2. He passes a bird at 1000 m1000 \ m of altitude at a speed of 100 m/s100\ m/s.
(a) How fast is he moving 1010 seconds later?