Inactive Tutor answered 05/15/23
distance traveled t=1 to t=3 is from -1 to -(1/9)
d=+8/9 length units for delta t=2 time units
vave=4/9 length units per time unit
d'(t)=v(t)=2t-3
v(1)=2 length units/time unit
v(3)=2/27 length units/time unit
Pam J.
asked 05/15/23If the function
over the time interval [1,3] defines the movement of a particle along the x-axis, then
A) Find the average velocity of the particle on the time interval, and
B) Find the instantaneous velocity of the particle at each endpoint of the time interval.
Inactive Tutor answered 05/15/23
distance traveled t=1 to t=3 is from -1 to -(1/9)
d=+8/9 length units for delta t=2 time units
vave=4/9 length units per time unit
d'(t)=v(t)=2t-3
v(1)=2 length units/time unit
v(3)=2/27 length units/time unit
Inactive Tutor answered 05/15/23
f(t) = -1/t^2
it helps to graph this, use a graphing calculator, handheld or on line
and you can see the slopes involved
f'(t) = 2/t^3
f'(1) = 2/1^3 = 2 = instantaneous rate of change at time t=1
f'(3) = 2/3^3 = 2/27 = instantaneous rate of charge at time t=3
(2-2/27)/(3-1) = (52/27)/2 = 26/27 = average rate of change from t=1 to 3
instantaneous rate of change = the slope of the tangent line
average rate of change = slope of the secant line
velocity = rate of change
Inactive Tutor
isn't (delta velocity)/(delta time)=average acceleration (units= distance/(time^2)? What if at both t=1 and t=3, that instananeous velocity is 2 ft./s. Does that mean the average velocity is zero?05/16/23
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Inactive Tutor
claim at end of answer is: ave. velocity is secant line (distance plotted as function of time) slope. Agreed. But then to compute ave. velocity you assume it is the difference between two instanteneous velocity values divided by respective change in time, which is not true. Ave. velocity is delta distance/delta time.05/15/23