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 hint is a dead givaway.

first : vf^2=vi^2 +2ad Vf=0, d=149.2, solve for a
then, solve for time
vf=vi+at solve for time t.
then,
df=di+vi*t+at^2
df=36.7 di=149.3 a same as above, vi=0
solve for time t.

Now add the times.

bobpursley, I did not understand your explanation :(

Could you please help? Thank you!!

I got two answers.

Using d=148.7, df=30.1, Vi=10m/s
First I got 26.73s, then I got 48.49s
not sure wich is right...if any

To find the time needed for the entire trip, we can break it down into two parts: the time until the object stops and the time it takes to travel from the displacement of 148.3 m to 36.7 m.

First, let's find the time until the object stops.

Since the object starts with an initial velocity of 10 m/s and comes to a stop, we know that the final velocity (vf) will be 0 m/s.

We also know that it undergoes a displacement of 148.3 m in the positive x direction. Let's call the time until it stops "t1".

We can use the equation of motion: vf = vi + at, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time.

In this case, the equation becomes: 0 = 10 + at1.

Now, let's find the time it takes to travel from the displacement of 148.3 m to 36.7 m.

The object starts from a displacement of 148.3 m and ends up with a displacement of 36.7 m. Let's call the time it takes to travel this distance "t2".

Since the object is still under the same acceleration, we can use the equation of motion: s = vit + (1/2)at^2, where s is the displacement, vi is the initial velocity, a is the acceleration, and t is the time.

In this case, the equation becomes: 36.7 = 0t2 + (1/2)a(t2)^2.

Now, we have two equations with two unknowns (t1 and t2). We can solve these equations simultaneously to find the values of t1 and t2.

To solve these equations, we can use the fact that the acceleration (a) is the same in both cases. Therefore, we can equate the two equations and solve for t2.

Setting up the equation: 10 + at1 = (1/2)a(t2)^2.

Divide both sides by a: 10/a + t1 = (1/2)(t2)^2.

Now, let's substitute the value of t1 from the first equation into the second equation and solve for t2.

Substituting 0 = 10 + at1, we get: 10/a + t1 = (1/2)(t2)^2.

Now, let's substitute the given values into the equation.

148.3 = 0(t1) + (1/2)a(t1)^2.

Now, we have two equations:

10/a + t1 = (1/2)(t2)^2, and
148.3 = 0(t1) + (1/2)a(t1)^2.

We can substitute this equation into the second equation.

148.3 = (1/2)a(t1)^2.

Now, we can solve for t1 by rearranging the equation.

Divide both sides by (1/2)a: (2 * 148.3) / a = (t1)^2.

Take the square root of both sides: sqrt((2 * 148.3) / a) = t1.

Now, we can substitute this value of t1 into the equation for t2:

10/a + sqrt((2 * 148.3) / a) = (1/2)(t2)^2.

Now, we can solve for t2 by rearranging the equation.

Subtract 10/a from both sides: sqrt((2 * 148.3) / a) = (1/2)(t2)^2 - 10/a.

Multiply both sides by 2: 2 * sqrt((2 * 148.3) / a) = (t2)^2 - 20/a.

Add 20/a to both sides: 2 * sqrt((2 * 148.3) / a) + 20/a = (t2)^2.

Take the square root of both sides: sqrt(2 * sqrt((2 * 148.3) / a) + 20/a) = t2.

Now that we have the values for t1 and t2, we can sum them to find the total time for the entire trip:

Total time = t1 + t2.