How do you calculate total distance?

Question

a particle moving at 10m/s reverses its direction to move at 10m/s in the opposite direction. if its acceleration is -10m/s2, what is the total distance that it travels?

a.0m

b.5m

c.10m

d.20m

i am clueless on this question and am overcaffeinated...any insight would be helpful..c is supposedly the correct answer.

Answer 1

This is kind of like the other question of yours that I just answered :)  Once again, you have to separate the forward motion and the reverse motion.  The forward motion is from v = 10 m/s to v = 0 with a = -10 m/s/s.  So,

v^2 = v0^2 + 2 a d

0 = (10)^2 + 2 (-10) d

--> d = 5m

The reverse motion is exactly identical, going from 0 to -10 m/s which gives another 5m in the opposite direction.  Hence the total distance traveled is 10m.

Answer 2

Are you sure c is correct?  It should be a.

v^2=u^2+2as => s=0

Answer 3

Consider initial direction as positive.

Initial velocity u = 10 m/s

Final velocity v = -10 m/s

Acceleration a = -10 m/s^2

v^2 = u^2 + 2as

Or, s = (v^2 - u^2)/2a

You are supposed to find total distance. The particle changes its speed from 10m/s to 0m/s and then it reverses its direction and moves at 10m/s in opposite direction.

Find distance of two parts and add the distances.

For first part:

u = 10m/s, v = 0, a = -10 m/s^2

s = (0 - 10^2)/2*(-10) = -100/-20 = 100/20 m = 5 m

For second part

u = 0, v = -10 m/s, a = -10 m/s^2

s = {(-10)^2 - 0}/2*(-10) = 100/(-20) = -5 m

Negative sign is because of reverse direction. You have to find distance and not displacement. Therefore, take absolute value.

Distance in second part = 5 m

Total distance = distance in first part + distance in second part

= 5 m + 5 m = 10 m

Ans: c