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.)

Question

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.

Answer 1

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.

Answer 2

bobpursley, I did not understand your explanation :(
Could you please help? Thank you!!

Answer 3

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