Velocity and Acceleration

Definition of Velocity and Speed

In single variable calculus the velocity is defined as the derivative of the position function.  For vector calculus, we make the same definition.

 Definition of Velocity Let r(t) be a differentiable vector valued function representing the position vector of a particle at time t.  Then the velocity vector is the derivative of the position vector.                v(t)  =  r'(t)  =  x'(t)i + y'(t)j + z'(t)k

Example

Find the velocity vector v(t) if the position vector is

r(t)  =  3ti + 2t2j - sin t k

Solution

We just take the derivative

v(t)  =  3i + 4tj + cos t k

When we think of speed, we think of how fast we are going.  Speed should not be negative.  In one variable calculus, speed was the absolute value of the velocity.  For vector calculus, it is the magnitude of the velocity.

 Definition of Speed Let r(t) be a differentiable vector valued function representing the position of a particle.  Then the speed of the particle is the magnitude of the velocity vector.           Speed  =  || v(t) ||  =  || r'(t) ||

Example

Let

r(t)  = 3i + 2t j + cos t k

Find the speed after p/4 seconds.

Solution

We first find the velocity vector

v(t)  =  r'(t)  =  2 j - sin t k

We have

v(p/4)  =  2 j - /2 k

Its magnitude is the square root of the sum of the squares or

Speed  =  || v ||  =

In one variable calculus, we defined the acceleration of a particle as the second derivative of the position function.  Nothing changes for vector calculus.

 Definition of Acceleration Let r(t) be a twice differentiable vector valued function representing the position vector of a particle at time t.  Then the acceleration vector is the second derivative of the position vector.                a(t)  =  r''(t)  =  x''(t)i + y''(t)j + z''(t)k

Example

Find the velocity and acceleration of the position function

r(t)  = (2t - 2) i + (t2 + t + 1)j

when t  =  -1.  Then sketch the vectors.

Solution

The velocity vector is

v(t)  =  r'(t)  =  2i  + (2t + 1) j

Plugging in -1 for t gives

v(-1)  =  2i - j

Take another derivative to find the acceleration.

a(t)  =  v'(t)  =  2j

Below is a picture of the vectors.

Projectile Motion

Since the velocity and acceleration vectors are defined as first and second derivatives of the position vector, we can get back to the position vector by integrating.

Example

You are a anti-missile operator and have spotted a missile heading towards you at the position

re  =  1000i + 500j

with velocity

ve  =  -30i + 3j

You can fire your anti-missile at 100 meters per second.  At what angle should you fire it so that you intercept the missile.  Assume that gravity is the only force acting on the projectiles.

Solution

The acceleration vector of the enemy missile is

ae(t)  =  -9.8 j

Integrating, we get the velocity vector

ve(t)  =  v1 i + (v2 - 9.8t) j

Setting t  =  0 and using the initial velocity of the enemy missile gives

ve(t)  =  -30 i + (3 - 9.8t) j

Now integrate again to find the position function

re(t)  =  (-30t + r1) i + (-4.9t2 + 3t + r2) j

Again setting t  =  0 and using the initial conditions gives

re(t)  =  (-30t + 1000) i + (-4.9t2 + 3t + 500) j

The acceleration of your anti-missile-missile is also

ay(t)  =  -9.8 j

Integrating, we get the velocity vector

vy(t)  =  v1 i + (v2 - 9.8t) j

Since the magnitude of our velocity is 100, we can say

vy(0)  =  100 cos q i + 100 sin q j

So that

vy(t)  =  100 cos q i + (100 sin q - 9.8t) j

Now integrate again to find the position function

ry(t)  =  (100t cos q + r1) i + (-4.9t2 + 100t sin q + r2) j

Our anti-missile-missile starts out at base, so the initial position is the origin.  All the constants are zero.

ry(t)  =  (100t cos q) i + (-4.9t2 + 100t sin q) j

Since we want to intercept the enemy missile, we set the position vectors equal to each other.

(100t cos q) i + (-4.9t2 + 100t sin q) j  =  (-30t + 1000) i + (-4.9t2 + 3t + 500) j

Equating coefficients gives

100t cos q =  -30t + 1000

-4.9t2 + 100t sin q  =  -4.9t2 + 3t + 500

The first equation gives

1000
t  =
100cos q  +  30

Simplifying the second equation and substituting gives

100000 sin q                    3000
=                                  +  500
100cos q  +  30          100cos q +  30

Clear denominators to get

100000 sin q  =  3000 + 50000 cos q + 15000

At this point we use a calculator to solve for q to