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.
Are my answers correct:
1) velocity function: 2t-6 and acceleration function: 2
2a) for position: s(1)=12-6(1)+5 = 0
b) s(2)=2^2-6(2)+5 = -3
c) s(3)=3^2-6(3)+5 = -4
d) s(5)=5^2-6(5)+5 = 0
e) s(8)=8^2-6(8)+5 = 21
2a) for velocity: 2(1)-6= -4
b) 2(2)-6= -2
c) 2(3)-6= 0
d) 2(5)-6= 4
e) 2(8)-6= 12
3) the acceleration is constant at 2.
all look ok to me.
good work.
1) To find the cyclist's velocity at any time t, you need to find the derivative of the position function s(t). The derivative of t^2 - 6t + 5 with respect to t is 2t - 6. Therefore, the velocity function is v(t) = 2t - 6.
Similarly, to find the cyclist's acceleration at any time t, you need to find the derivative of the velocity function v(t). The derivative of 2t - 6 with respect to t is 2. Therefore, the acceleration function is a(t) = 2.
2) Let's find the position and velocity of the cyclist after different time intervals:
a) After one hour (t = 1):
- Position: Plug t = 1 into the position function s(t) = t^2 - 6t + 5.
s(1) = (1)^2 - 6(1) + 5 = 0
Interpretation: The cyclist is at 0 miles north of Los Angeles after one hour.
- Velocity: Plug t = 1 into the velocity function v(t) = 2t - 6.
v(1) = 2(1) - 6 = -4
Interpretation: The cyclist has a velocity of -4 miles per hour after one hour. This means the cyclist is traveling southward.
b) After two hours (t = 2):
- Position: Plug t = 2 into the position function s(t).
s(2) = (2)^2 - 6(2) + 5 = -3
Interpretation: The cyclist is at -3 miles north of Los Angeles after two hours.
- Velocity: Plug t = 2 into the velocity function v(t).
v(2) = 2(2) - 6 = -2
Interpretation: The cyclist has a velocity of -2 miles per hour after two hours.
c) After three hours (t = 3):
- Position: Plug t = 3 into the position function s(t).
s(3) = (3)^2 - 6(3) + 5 = -4
Interpretation: The cyclist is at -4 miles north of Los Angeles after three hours.
- Velocity: Plug t = 3 into the velocity function v(t).
v(3) = 2(3) - 6 = 0
Interpretation: The cyclist has a velocity of 0 miles per hour after three hours. This means the cyclist has stopped or is momentarily at rest.
d) After five hours (t = 5):
- Position: Plug t = 5 into the position function s(t).
s(5) = (5)^2 - 6(5) + 5 = 0
Interpretation: The cyclist is at 0 miles north of Los Angeles after five hours.
- Velocity: Plug t = 5 into the velocity function v(t).
v(5) = 2(5) - 6 = 4
Interpretation: The cyclist has a velocity of 4 miles per hour after five hours.
e) After eight hours (t = 8):
- Position: Plug t = 8 into the position function s(t).
s(8) = (8)^2 - 6(8) + 5 = 21
Interpretation: The cyclist is at 21 miles north of Los Angeles after eight hours.
- Velocity: Plug t = 8 into the velocity function v(t).
v(8) = 2(8) - 6 = 12
Interpretation: The cyclist has a velocity of 12 miles per hour after eight hours.
3) The acceleration of the cyclist is constant and equal to 2 for all times. This means the cyclist is experiencing a constant acceleration of 2 miles per hour squared. It indicates that the cyclist's velocity is increasing by 2 miles per hour every hour.