Calculating Total Distance Traveled with Constant Acceleration

  • Thread starter Joe26
  • Start date
In summary, to find the total distance traveled by an object with constant acceleration, you must break the problem into two parts - the portion of the motion in the negative direction and the portion in the positive direction. Find the time it takes for the object to reach its turning point and use the displacement formula to find the distance traveled in each direction. Then add these distances together to get the total distance traveled.
  • #1
Joe26
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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?
 
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  • #2
Welcome to PF Joe26,

Joe26 said:

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.
 
  • #3
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!
 

Related to Calculating Total Distance Traveled with Constant Acceleration

1. What is total distance?

Total distance is the sum of all distances traveled during a specific period of time or over a specific route.

2. How do you calculate total distance?

The total distance can be calculated by adding up all the individual distances traveled. This can be done by using a formula or by simply adding up the distances manually.

3. What is the difference between total distance and displacement?

Total distance takes into account all the distances traveled, regardless of direction, while displacement only considers the straight-line distance between the starting and ending points.

4. Can total distance be negative?

No, total distance cannot be negative. Distance is a scalar quantity and only has a magnitude, not a direction. Therefore, it is always positive or zero.

5. How does total distance differ from total displacement?

Total distance is the sum of all distances traveled, while total displacement is the difference between the starting and ending points. Total displacement can be smaller than or equal to the total distance, but it cannot be greater.

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