2.5: Velocity and Acceleration
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In single variable calculus the velocity is defined as the derivative of the position function. For vector calculus, we make the same definition.
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)ˆi+y′(t)ˆj+z′(t)ˆk.
Find the velocity vector v(t) if the position vector is
r(t)=3tˆi+2t2ˆj+sin(t)ˆk.
Solution
We just take the derivative
v(t)=3ˆi+4tˆj+cos(t)ˆk.
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.
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)||.
Let
r(t)=3ˆi+2ˆj+costˆk.
Find the speed after p4 seconds.
Solution
We first find the velocity vector
v(t)=r′(t)=2ˆj−sin(t)ˆk.
We have
v(p4)=2ˆj−√22.
Its magnitude is the square root of the sum of the squares or
speed=||v||=√22+(√22)2=√4.5.
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.
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)ˆi+y″(t)ˆj+z″(t)ˆk
Find the velocity and acceleration of the position function
r(t)=(2t−2)ˆi+(t2+t+1)ˆj
when t=−1. Then sketch the vectors.
Solution
The velocity vector is
v(t)=r′(t)=2ˆi+(2t+1)ˆj.
Plugging in -1 for t gives
b(−1)=2ˆi−ˆj.
Take another derivative to find the acceleration.
a(t)=v′(t)=2ˆj.
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.
You are a anti-missile operator and have spotted a missile heading towards you at the position
re=1000ˆi+500ˆj
with velocity
ve=−30ˆi+3ˆj.
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)=−9.8ˆj.
Integrating, we get the velocity vector
ve(t)=v1ˆi+(v2−9.8t)ˆj.
Setting t=0 and using the initial velocity of the enemy missile gives
ve(t)=−30ˆi+(3−9.8t)ˆj.
Now integrate again to find the position function
re(t)=(−30t+r1)ˆi+(−4.9t2+3t+r2)ˆj.
Again setting t=0 and using the initial conditions gives
re(t)=(−30t+1000)ˆi+(−4.9t2+3t+500)ˆj.
The acceleration of your anti-missile-missile is also
ay(t)=−9.8tˆj.
Integrating, we get the velocity vector
vy(t)=v1ˆi+(v2−9.8t)ˆj.
Since the magnitude of our velocity is 100, we can say
vy(0)=100cosqˆi+100sinqˆj.
So that
vy(t)=100cosqˆi+(100sinq−9.8t)ˆj.
Now integrate again to find the position function
ry(t)=(100tcosq+r1)ˆi+(−4.9t2100sinq−9.8t+r2)ˆj.
Our anti-missile-missile starts out at base, so the initial position is the origin. All the constants are zero.
ry(t)=(100tcosq)ˆi+(−4.9t2100sinq−9.8t)ˆj
Since we want to intercept the enemy missile, we set the position vectors equal to each other.
(100tcosq)ˆi+(−4.9t2100sinq−9.8t)ˆj=(−30t+1000)ˆi+(−4.9t2+3t+500)ˆj
Equating coefficients gives
100tcosq=−30t+1000
−4.9t2+100tsinq=−4.9t2+3t+500.
The first equation gives
t=1000100cosq+30.
Simplifying the second equation and substituting gives
100000sinq100cosq+30=3000100cosq+30+500.
Clear denominators to get
100000sinq=3000+50000cosq+15000.
At this point we use a calculator to solve for q to
q=0.62535rads.