A car travels in a straight line

  • #1
a car travels in a straight line....

Homework Statement



A car travels in a straight line for 4.8 h at a
constant speed of 89 km/h.
What is its acceleration?
Answer in units of m/s2

Homework Equations



here is what i'm confused about: I only have time and speed, how exactly could i apply the acceleration formula if i dont have the initial or final velocity?


The Attempt at a Solution



was confused to attempt :confused::confused:
 

Answers and Replies

  • #2
lewando
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Well, you do! If speed is constant for that time interval, the v0 = vf = 89 km/h.
 
  • #3


sorry bear with me.....so my initial velocity and final velicty is zero? :confused:
 
  • #4
lewando
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No, I don't read that. I read 89 km/h.
 
  • #5


but doesnt constant speed mean no acceleration? cause its moving in a straight line
 
  • #6
lewando
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We have no knowledge of the velocity before or after this time interval. So stay inside the time interval. [yes constant speed [velocity!] means 0 acceleration]
 
  • #7


First you need to convert the two values. To get km/h to m/s you multiply by 1000 and divide by 3600. To get 4h to seconds you multiply 4 by 3600

89 km/h * 1000 / 3600 = 24.73 m/s, 4h = 14400sec

The acceleration equation is:

accel = v2-v1/t2-t1

v2 = 24.73 m/s, v1 = 24.73 m/s (because it's constant), t2 = 14400 sec, t1 = 0 sec

So:

accel = 24.73 - 24.73 / 14400 - 0 = 0 m/s^2

The acceleration is 0 because the velocity is constant over the time frame. Since acceleration is the slope of velocity, there has to be a change in the velocity for there to be a non-zero for acceleration.
 
  • #8


question: where did you get the formula v2-v1/t2-t1? because i dont think we've gotten that far yet in class. we just got: vf-vi/t .

so the acceleration is 0. ok thanks to both of u who helped. :)
 
  • #9


v2-v1/t2-t1 is just the slope formula y2-y1/x2-x1

Since acceleration is the slope (or the derivative) of velocity, finding the slope between two points of velocity will give you average acceleration.

Your equation is pretty much the exact same as that, only your t doesn't bother with subtracting the initial time since the initial time is 0. vf = 24.73 m/s, vi = 24.73 m/s and t = 14400 sec.
 
  • #10


one more thing sorry to be a bother on this question but if it says its constant speed, does it mean the accererlation is always 0?
 
  • #11
lewando
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The original question says "constant speed" and "straight line". This implies constant velocity. Constant velocity implies zero acceleration.
 
  • #12


Yes, because speed is the magnitude of the velocity. If the magnitude of the velocity never changes, then the acceleration is 0 m/s^2.
 
  • #13


so like for this question:
A rocket-driven sled running on a straight,
level track has been used to study the physio-
logical effects of large accelerations on astro-
nauts. One such sled can attain a speed of
397 m/s in 1.6 s starting from rest.
What is the acceleration of the sled, assum-
ing it is constant?

since the rocket thing is going straight, and its assuming its constant, the acceleration would be 0 right? or am I getting it all wrong?

sorry i just got this hw from physics and its like some type of online thing where if you put in the answer just the very first time, and its wrong, you lose like 20 points. i know it sucks :(. just im confused on some things.
 
  • #14
lewando
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so like for this question:
A rocket-driven sled running on a straight,
level track has been used to study the physio-
logical effects of large accelerations on astro-
nauts. One such sled can attain a speed of
397 m/s in 1.6 s starting from rest.
What is the acceleration of the sled, assum-
ing it is constant?

since the rocket thing is going straight, and its assuming its constant, the acceleration would be 0 right? or am I getting it all wrong?

sorry i just got this hw from physics and its like some type of online thing where if you put in the answer just the very first time, and its wrong, you lose like 20 points. i know it sucks :(. just im confused on some things.

In this case, the horizontal acceleration is NOT zero. Standard kinematic equations with constant acceleration apply.
 
  • #15


The acceleration is constant, but it is not zero. If it was 0, it would not be possible for the sled to get to a velocity of 397 m/s in 1.6s from rest (0 m/s).

Trying using the equation V = Vo + a(t) and solve for a.
 
  • #16


ok but does that mean my initial velocity is 397 or is that my final?
 
  • #17


397 m/s would be your final velocity.
 
  • #18


ok i think i got it,
i plugged in the numbers into that equation and here is what i got:
V=Vo + a(t)
V=397
Vo=0
a=?
t=1.6

397=0+a(1.6)
397=a(1.6) *divide by 1.6
getting :
a=248.13 m/s^2


idk....it says this question next to it saying: How far does the sled travel in 1.6 s, starting
from rest? but isnt that the same question in the first place? is there an error or something?
 
  • #19


The next question is asking for the displacement (or how far the sled traveled in meters) of the sled after 1.6 seconds. The sled couldn't have stayed in place if it's velocity increased to 397 m/s after 1.6seconds, so it has to have a displacement.

Trying using the equation: deltaX = Vo(t) + 1/2(a)(t^2)
 
  • #20


oh, so then using that equation i get:

deltaX = Vo + 1/2(a)(t^2)

deltaX=0+1/2(248.13)(1.6^2)
deltaX=317.61 m

so far so good?
 
  • #21


That's what I got as my answer to the problem.
 
  • #22


ok. well thankyou fisherman166 for all your help! even though i asked many questions and was being annoying lol :redface::redface: but i understand some concepts i didn't understand now. :approve: :D :D
 
  • #23


My pleasure! I'm glad I could help out. This is also my first week of physics with calculus, so this was a good exercise to make sure I could use my constant acceleration equations correctly :D
 

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