# Answer to Question #15873 in Mechanics | Relativity for Dp

Question #15873

Moving man starts in the negative direction with a speed of 3 m/s but is accelerating in the positive direction with acceleration of 2.75m/s^2. What is his velocity and displacement at the end of 5.00 seconds?

Expert's answer

The man's velocity at the end of 5.00 seconds is

V = -3[m/s] + 2.75[m/s²]*5.00[s] = 10.75[m/s].

After

T = 3[m/s]/2.75[m/s²] = 12/11 seconds

the man will stop. After that he begin to run in the positive direction. So, his displacement is

D = 2.75[m/s²]*(5.00[s]-12/11[s])²/2 ≈ 21 metres

in positive direction.

V = -3[m/s] + 2.75[m/s²]*5.00[s] = 10.75[m/s].

After

T = 3[m/s]/2.75[m/s²] = 12/11 seconds

the man will stop. After that he begin to run in the positive direction. So, his displacement is

D = 2.75[m/s²]*(5.00[s]-12/11[s])²/2 ≈ 21 metres

in positive direction.

Need a fast expert's response?

Submit orderand get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

## Comments

## Leave a comment