Calculating Total Distance Traveled with Constant Acceleration

[Joe26]

Homework Statement

An object moves with constant acceleration 4.60 m/s2 and over a time interval reaches a final velocity of 10.0 m/s. If its initial velocity is −5 m/s, what is its displacement during the time interval? What is the total distance travelled?

a=4.60m/s^2
vi=-5 m/s
vf=10m/s

Homework Equations

vf= (vi)+(a)(t)
deltax=(vi)(t) + 1/2(a)(t)^2

The Attempt at a Solution

vf=vi+at
10=-5+(4.6)t
t=3.26 s

deltax= vit+1/2(a)(t)^2
deltax= (-5)(3.26)+1/2(4.6)(3.26)^2
deltax= 8.14 m (displacement)

total distance traveled = 8.14 m

the answer i got for displacement is correct but the total distance traveled is wrong and i can't figure out why. can anyone help me please?

 

[cepheid]

Welcome to PF Joe26,

Homework Statement

An object moves with constant acceleration 4.60 m/s2 and over a time interval reaches a final velocity of 10.0 m/s. If its initial velocity is −5 m/s, what is its displacement during the time interval? What is the total distance travelled?

a=4.60m/s^2
vi=-5 m/s
vf=10m/s

Homework Equations

vf= (vi)+(a)(t)
deltax=(vi)(t) + 1/2(a)(t)^2

The Attempt at a Solution

vf=vi+at
10=-5+(4.6)t
t=3.26 s

deltax= vit+1/2(a)(t)^2
deltax= (-5)(3.26)+1/2(4.6)(3.26)^2
deltax= 8.14 m (displacement)

total distance traveled = 8.14 m

the answer i got for displacement is correct but the total distance traveled is wrong and i can't figure out why. can anyone help me please?

The object is initially traveling in one direction, but then it slows, stops, and begins traveling in the opposite direction. If you just use that equation above for delta x, the movement in the negative direction will cancel out part of the movement in the positive direction, and you'll just end up with the net change in position (i.e. the displacement), which is how far it ended up from its starting point (when all was said and done).

What you need to do is figure out how far it went in the negative direction, and then how far it traveled in the positive direction after its turn-around. Then add these together to get the total distance travelled, which will be larger than the net displacement. So, you sort of have to break the problem into two parts: the portion during which the motion was in the negative direction, and the portion during which the motion was in the positive direction.

 

[Joe26]

thank you for your help, i split the problem into two parts and got my answer! what i did first was to find the time it takes it get from velocity=-5m/s to velocity=0m/s (vi=-5m/s; vf=0m/s). Then i used the displacement formula to find distance in the negative direction. After that, i found the time it takes to get to from velocity=0m/s to velocity=10m/s (vi=0m/s; vf=10m/s). Then i used the displacement formula to find distance in the positive direction. Added the two distances to get total distance travelled. Thanks for your help!