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
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
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
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.
1 answer