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)?

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.

To solve these questions, we need to use the equation for velocity and apply it to each part of the problem. Let's go step by step:

a. To find the time it takes for the object to come to a stop (velocity = 0), we need to set the velocity equation to 0 and solve for t.

v(t) = 38 - 2t
0 = 38 - 2t

To solve for t, we can subtract 38 from both sides:

-38 = -2t

Then divide both sides by -2:

t = -38 / -2
t = 19

So, it takes 19 seconds for the object to come to a stop.

b. To find how far the object travels during that time, we can use the formula for displacement. Displacement equals the integral of velocity with respect to time from 0 to t.

d(t) = ∫v(t) dt (from 0 to t)
d(t) = ∫(38 - 2t) dt (from 0 to t)

Integrating the function gives:

d(t) = [38t - t^2] from 0 to t

Plugging in the value of t:

d(t) = (38t - t^2) - (38(0) - (0)^2)
d(t) = 38t - t^2

So, the distance traveled during that time is given by the function d(t) = 38t - t^2.

c. To find the time it takes for the car to travel half the distance found in part b, we need to set d(t) equal to half of the distance traveled.

d(t) = 38t - t^2
1/2(38t - t^2) = 19t - (t^2)/2

Simplifying the equation gives:

19t - (t^2)/2 = 19t

Subtracting 19t from both sides:

-(t^2)/2 = 0

To solve for t, we set the equation equal to zero:

t^2 = 0

The only solution to this equation is t = 0 since any number multiplied by zero is zero.

Therefore, it would take zero seconds for the car to travel half of the distance found in part b.

Note: The fact that t=0 is the only solution indicates that the car reaches half the distance instantaneously. This happens because the equation for displacement is a quadratic equation, and one of its roots is 0, which represents the starting point of the car.