An object's velocity after t seconds is v(t)= 38-2t feet per second.

a. How many seconds does it take for the object to come to a stop (velocity = 0)?

b. How far does the car travel during that time?

c. How many seconds does it take the car to travel half the distance in part (b)?

8 answers

I do not know if you know calculus or if you have kinetatic equations in your text. What we have here is a constant acceleration of -2 ft/s^2

a. 0 = 38 - 2 t
2 t = 38
t = 19 s

b. d = Vi t + (1/2) a t^2
= 38 (19) - 19^2
= 19(19) = 361 feet
or use average speed for 19 seconds
= (38/2)*19

c. 180.5 = 38 t - t^2
t^2 - 38 t + 180.5 = 0
t = [38+/-sqrt(1444-722)]/2
= [38 +/- 26.9]/2
use the answer that is less than 19 seconds
= 5.56 s
its calculus but will kinetatic equations work as well?
oh
well
v(t)= 38-2t
or
dx/dt = 38 - 2 t
then
x = c + 38 t - (1/2)(2)t^2
c = x when t = 0
call it 0 here
x = 38 t - t^2
I used the kinematics because I did not know if you knew how to integrate velocity to get distance.
kinematics look easier i can do both ways but will they have different values
No, same answers both ways.
Thanks Damon!
You are welcome.
Similar Questions
  1. Use the velocity vs. time graph to answer the question.At which two moments would the acceleration of the object be the same? (1
    1. answers icon 1 answer
    1. answers icon 3 answers
  2. An object's velocity after t seconds is v(t)= 38-2t feet per second.a. How many seconds does it take for the object to come to a
    1. answers icon 0 answers
  3. The graph shows an object in motion for 14 seconds.At which time interval on the graph is the object moving at a constant
    1. answers icon 1 answer
more similar questions