s(t) = t^2 - 6t + 5 models the motion of a person cycling along Rte 66 where s(t) is the number of miles north of Los Angeles the person is at time t hours.
1) Write functions for the cyclist's velocity and acceleration at any time t.
2) Find the position and velocity of the cyclist after: (Be sure to interpret your findings in terms of the problem.)
a) one hour
b) two hours
c) three hours
d) five hours
e) eight hours
3) What is the acceleration of the cyclist after each time? Explain these results in terms of the problem.
since you are taking Calculus, you MUST know that velocity is the derivative of distance, and acceleration is the derivative of velocity
1. so take those derivatives
2. a) plug in t = 1 into both your distance and your velocity expression
etc
3. That should be a constant. After all if you differentiate a quadratic term twice you end up with a constant
i.e. t^2 ---> 2t ---> 2
s(t)=t²-6t+5 is the position vector.
For example, at time=0 hour,
s(0)=5, so the person is 5 miles north of LA, etc.
Velocity is the derivative of position with respect to time, namely,
v(t)=s'(t)=2t-6
At time t=0,
v(t)=-6, meaning he was cycling towards LA.
Acceleration is the derivative of velocity with respect to time, so
a(t)=v'(t)=2
At all times, acceleration is positive.
Will leave part (2) for you to practice on (interpretation).
1) To find the cyclist's velocity function, we need to take the derivative of the position function, s(t), with respect to time, t. The derivative of t^2 is 2t, the derivative of -6t is -6, and the derivative of 5 (a constant) is zero. Therefore, the velocity function, v(t), is given by:
v(t) = 2t - 6
To find the cyclist's acceleration function, we need to take the derivative of the velocity function, v(t), with respect to time, t. The derivative of 2t is 2, and the derivative of -6 (a constant) is zero. Therefore, the acceleration function, a(t), is given by:
a(t) = 2
2)
a) To find the position and velocity of the cyclist after one hour, we substitute t = 1 into the position function and the velocity function:
s(1) = 1^2 - 6(1) + 5 = 0 miles north of Los Angeles
v(1) = 2(1) - 6 = -4 miles per hour
Interpretation: After one hour, the cyclist is at position s(1) = 0 miles north of Los Angeles, indicating that they have reached Los Angeles. The negative velocity v(1) = -4 miles per hour indicates that the cyclist is traveling southward at a speed of 4 miles per hour.
b) To find the position and velocity of the cyclist after two hours, we substitute t = 2 into the position function and the velocity function:
s(2) = 2^2 - 6(2) + 5 = -1 mile north of Los Angeles
v(2) = 2(2) - 6 = -2 miles per hour
Interpretation: After two hours, the cyclist is at position s(2) = -1 mile north of Los Angeles, indicating that they are 1 mile south of Los Angeles. The negative velocity v(2) = -2 miles per hour indicates that the cyclist is still traveling southward but at a slower speed of 2 miles per hour.
c) To find the position and velocity of the cyclist after three hours, we substitute t = 3 into the position function and the velocity function:
s(3) = 3^2 - 6(3) + 5 = 2 miles north of Los Angeles
v(3) = 2(3) - 6 = 0 miles per hour
Interpretation: After three hours, the cyclist is at position s(3) = 2 miles north of Los Angeles, indicating that they have started moving north again. The velocity v(3) = 0 miles per hour indicates that the cyclist has momentarily stopped.
d) To find the position and velocity of the cyclist after five hours, we substitute t = 5 into the position function and the velocity function:
s(5) = 5^2 - 6(5) + 5 = -5 miles north of Los Angeles
v(5) = 2(5) - 6 = 4 miles per hour
Interpretation: After five hours, the cyclist is at position s(5) = -5 miles north of Los Angeles, indicating that they are 5 miles south of Los Angeles. The positive velocity v(5) = 4 miles per hour indicates that the cyclist is traveling northward at a speed of 4 miles per hour.
e) To find the position and velocity of the cyclist after eight hours, we substitute t = 8 into the position function and the velocity function:
s(8) = 8^2 - 6(8) + 5 = 21 miles north of Los Angeles
v(8) = 2(8) - 6 = 10 miles per hour
Interpretation: After eight hours, the cyclist is at position s(8) = 21 miles north of Los Angeles. The positive velocity v(8) = 10 miles per hour indicates that the cyclist is still traveling northward but at a faster speed of 10 miles per hour.
3) The acceleration of the cyclist is a constant value of 2. This means that regardless of the time, the cyclist's acceleration remains unchanged. In terms of the problem, it indicates that the cyclist's speed is increasing by 2 miles per hour every hour. The acceleration being constant suggests that there are no external forces acting on the cyclist to affect their acceleration.