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 ||  = 

Acceleration

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

        r_e  =  1000i + 500j

with velocity 

          v_e  =  -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 

        a_e(t)  =  -9.8 j 

Integrating, we get the velocity vector

        v_e(t)  =  v₁ i + (v₂ - 9.8t) j

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

        v_e(t)  =  -30 i + (3 - 9.8t) j

Now integrate again to find the position function

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

Again setting t  =  0 and using the initial conditions gives

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

The acceleration of your anti-missile-missile is also

        a_y(t)  =  -9.8 j 

Integrating, we get the velocity vector

        v_y(t)  =  v₁ i + (v₂ - 9.8t) j

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

        v_y(0)  =  100 cos q i + 100 sin q j

So that

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

Now integrate again to find the position function

        r_y(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.

        r_y(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

        q  =  .62535 radians  

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