An object has a initial velocity of 10 m/s in the positive x direction. Under a constant acceleration it stops after undergoing a displacement of 148.3 m in the positive x direction. Because it is still under the same acceleration it then travels back in the negative in the x direction and ends up with a displacement of 36.7 m in the positive x direction relative to where it started. What is the time needed for the entire trip in seconds? (From the start until it arrives back where the displacement is 36.7 m in the positive x direction.)

Hint: It's easier to do this in two parts. First find the time until it stops, and then, in the second part, find the time it takes to get from the displacement of 148.3 m until 36.7 m in the positive x direction. Then sum the two times.

The first phase is easy if you have constant acceleration because the average speed until the stop is Vi/2 = 5 m/s

t to stop = 148.3 / 5 = 29.7 seconds

a = change in velocity/change in time = -10/29.7 = -.337 m/s^2

phase 2 displacement = 37.7 - 148.3
= -110.6 meters

-110.6 = (1/2)(-.337)t^2
solve for t
add that t to 29.7 seconds to get the total time

To find the total time needed for the entire trip, we can break it down into two parts: the time it takes for the object to come to a stop and the time it takes for the object to travel from a displacement of 148.3 m to 36.7 m in the positive x direction.

First, let's determine the time it takes for the object to come to a stop. We are given the initial velocity (10 m/s), the displacement (148.3 m), and the fact that it is under a constant acceleration.

We can use the following kinematic equation to solve for the time (t1) it takes for the object to come to a stop:

v^2 = u^2 + 2as

Where:
v = final velocity (0 m/s since it comes to a stop)
u = initial velocity (10 m/s)
a = acceleration (constant, but unknown)
s = displacement (148.3 m)

Rearranging the equation, we get:

0 = (10 m/s)^2 + 2a(148.3 m)

Simplifying the equation:

0 = 100 m^2/s^2 + 296.6a

Now, let's solve for the acceleration (a):

296.6a = -100 m^2/s^2

a = -100 m^2/s^2 / 296.6

a ≈ -0.34 m/s^2

Now that we have the acceleration, we can use the first kinematic equation again (v = u + at) to find the time (t1) it takes for the object to come to a stop:

0 m/s = 10 m/s + (-0.34 m/s^2) * t1

Simplifying the equation:

-10 m/s = -0.34 m/s^2 * t1

t1 = -10 m/s / (-0.34 m/s^2)

t1 ≈ 29.41 s

So, it takes approximately 29.41 seconds for the object to come to a stop.

Now let's move on to the second part: the time it takes for the object to travel from a displacement of 148.3 m to 36.7 m in the positive x direction.

To find the time (t2) it takes for this part of the trip, we can use the second kinematic equation (s = ut + (1/2)at^2), where u is the final velocity from the previous part (0 m/s), s is the displacement (36.7 m), and a is the same acceleration (-0.34 m/s^2).

36.7 m = 0 m/s * t2 + (1/2)(-0.34 m/s^2)(t2)^2

Simplifying the equation:

36.7 m = (1/2)(-0.34 m/s^2)(t2)^2

73.4 m = (-0.34 m/s^2)(t2)^2

Now, solve for t2:

(t2)^2 = 73.4 m / (-0.34 m/s^2)

t2 ≈ √(-73.4 m / -0.34 m/s^2)

t2 ≈ √215.882 s^2

t2 ≈ 14.70 s

So, it takes approximately 14.70 seconds for the object to travel from a displacement of 148.3 m to 36.7 m in the positive x direction.

To find the total time for the entire trip, we sum up t1 and t2:

Total time = t1 + t2
≈ 29.41 s + 14.70 s
≈ 44.11 s

Therefore, the total time needed for the entire trip is approximately 44.11 seconds.