A particle moves along a line so that its posistion at any t is greater than or equal to 0 is given by the function s(t)= t^3-8t+1, where s is measured in feet and t is measured in seconds.
a) find the displacement during the first three seconds
b) Find the average velocity during the first three seconds
c) Find the instantaneous velocity when t=3 seconds
d) find the accelteration of the object when t=3 seconds
e) At what value or values of t does the particle change direction?
a) find s(3) - s(0)
b) avg vel in first 3 sec = (s(3) - s(0))/3
c) find derivative of s(t), then sub in t = 3
d) find derivative of c) result, sub in t=3
e) mmmh, one of those derivatives (velocity or acceleration ) must be zero when I change direction ...
a) To find the displacement during the first three seconds, we need to find the difference between the position at 3 seconds and the position at 0 seconds.
s(3) = (3)^3 - 8(3) + 1
s(3) = 27 - 24 + 1
s(3) = 4 feet
s(0) = (0)^3 - 8(0) + 1
s(0) = 0 - 0 + 1
s(0) = 1 feet
Displacement = s(3) - s(0)
Displacement = 4 - 1
Displacement = 3 feet
b) To find the average velocity during the first three seconds, we need to divide the displacement by the time interval. In this case, the time interval is 3 seconds.
Average velocity = Displacement / Time interval
Average velocity = 3 feet / 3 seconds
Average velocity = 1 foot/second
c) To find the instantaneous velocity when t = 3 seconds, we need to find the derivative of the position function with respect to time and evaluate it at t = 3.
v(t) = ds(t) / dt
v(t) = 3t^2 - 8
v(3) = 3(3)^2 - 8
v(3) = 27 - 8
v(3) = 19 feet/second
The instantaneous velocity when t = 3 seconds is 19 feet/second.
d) To find the acceleration of the object when t = 3 seconds, we need to find the derivative of the velocity function with respect to time and evaluate it at t = 3.
a(t) = dv(t) / dt
a(t) = 6t
a(3) = 6(3)
a(3) = 18 feet/second^2
The acceleration of the object when t = 3 seconds is 18 feet/second^2.
e) The particle changes direction when its velocity changes sign. To find the values of t where this occurs, we need to find the zeros of the velocity function.
3t^2 - 8 = 0
t^2 = 8/3
t = ±√(8/3)
So, the particle changes direction at t = √(8/3) and t = -√(8/3) seconds.
To solve these problems, we'll need to use calculus. I'll walk you through each step to find the answers.
a) The displacement during the first three seconds can be found by calculating the difference in position between t = 3 and t = 0. So, substitute t = 3 into the equation for position:
s(3) = (3^3) - (8 * 3) + 1
= 27 - 24 + 1
= 4 feet
Therefore, the displacement during the first three seconds is 4 feet.
b) The average velocity during the first three seconds can be found by calculating the ratio of the displacement to the time taken. In this case, the time taken is 3 seconds, and we already found the displacement to be 4 feet. So:
Average velocity = displacement / time taken
= 4 feet / 3 seconds
Hence, the average velocity during the first three seconds is 4/3 feet per second.
c) To find the instantaneous velocity when t = 3 seconds, we need to find the derivative of the position function s(t) with respect to time. So, take the derivative of s(t):
s'(t) = 3t^2 - 8
Evaluate this derivative at t = 3:
s'(3) = 3(3^2) - 8
= 3(9) - 8
= 27 - 8
= 19 feet per second
Therefore, the instantaneous velocity when t = 3 seconds is 19 feet per second.
d) The acceleration is found by taking the derivative of the velocity function or the second derivative of the position function. Since we already have the velocity function from part c), let's find the second derivative:
s''(t) = d^2(s)/dt^2 = d(s'(t))/dt
Take the derivative of s'(t):
s''(t) = d/dt(3t^2 - 8)
= 6t
Evaluate this second derivative at t = 3:
s''(3) = 6(3)
= 18 feet per second squared
Thus, the acceleration of the object when t = 3 seconds is 18 feet per second squared.
e) To find the value or values of t when the particle changes direction, we need to locate the points where the velocity changes sign. These points occur when the velocity function, s'(t), is equal to zero:
3t^2 - 8 = 0
To solve for t, we can factor the equation:
3t^2 = 8
t^2 = 8/3
t = ±√(8/3)
So, the particle changes direction at t = √(8/3) and t = -√(8/3).
That's it! You have solved all parts of the problem using calculus.