Home > Calculus > Net Change: Motion on a Line

How do you find the total displacement for the particle whose position at time #t# is given by #s(t)=3t^2-5t# on the interval #0<=t<=2#?

Answer 1

The total displacement will be the difference between the final and initial positions. You have: Initial position; for #t=0# #s_i=s(0)=3×0^2-5×0=0# Final position; for #t=2# #s_f)=s(2)=3×(2)^2-5×2=12-10=2#

Total displacement#=s_f)-s_i=2-0=2# (Km, m, cm, mm....)

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Ezra Adams

To find the total displacement of the particle on the interval (0 \leq t \leq 2), we need to calculate the change in position from the initial time to the final time. The total displacement is given by the difference in position at the final time and the initial time.

First, we find the position of the particle at (t = 0): [s(0) = 3(0)^2 - 5(0) = 0]

Next, we find the position of the particle at (t = 2): [s(2) = 3(2)^2 - 5(2) = 3(4) - 5(2) = 12 - 10 = 2]

The total displacement is then: [ \text{Total displacement} = s(2) - s(0) = 2 - 0 = 2 ]

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you find the total displacement for the particle whose position at time #t# is given by #s(t)=3t^2-5t# on the interval #0&lt;=t&lt;=2#?

How do you find the total displacement for the particle whose position at time #t# is given by #s(t)=3t^2-5t# on the interval #0<=t<=2#?