How Do You Calculate Acceleration and Distance in Physics Problems?

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Homework Help Overview

The problem involves a car's acceleration, distance traveled during acceleration, work done, and power developed over a specified time period. The subject area is kinematics and dynamics in physics.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculation of acceleration and distance, with initial confusion over unit conversions and the correct application of formulas. There is an exploration of different methods to calculate force and work, along with questions about the assumptions regarding horizontal versus vertical motion.

Discussion Status

Some participants have recalculated values and clarified unit conversions, leading to revised acceleration and distance results. There is ongoing dialogue about the appropriate formulas for calculating force and work, with some guidance provided on the context of horizontal motion.

Contextual Notes

Participants note the importance of unit consistency and the implications of motion direction on the choice of formulas. There is mention of checking answers against online resources, indicating a reliance on external validation for correctness.

ElegantSir
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Homework Statement


"A 1480 kg car accelerates uniformly from a position of rest to a speed of 95km/h in 6 s?"

1) calculate the cars acceleration
2) determine how far the car traveled during this acceleration
3) determine how much work was done by the car during the 6s
4) identify what power was developed during this process

Homework Equations


v2 = v1 + at
d = v1t + 1/2at²
d = v2 - 1/2at²
work = force / distance or w = f / d
power = work / time or p = w / t
f = m * g

The Attempt at a Solution


1) I rearranged v2 = v1 + at to a = v2 - v1 / t to find power
so
95km/h - 0km/h / 6s = 95km/h / 6s = 15.83m/s²
Therefore the acceleration of the car should be 15.83m/s²

2) This is where I am stuck.
I would assume I need to use the equation d = v1t + 1/2at²
so
d = 0 * 6 + 1/2 * 15.83 * (6)²
from this i get 284.94m[/B]

I usually check online for the answers after I am done my work but none of the online sources I am finding match my answers for question 2. The answer I am finding online is 79.167. Would anybody be able to tell me if both of the answers for question 1 and 2 are correct?
Any help is appreciated!
 
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You need to keep better track of your units. km/hr divided by s doesn't give m/s^2. Try converting the 95 km/hr into m/s.
 
Ahh I completely missed that. so 95km/h becomes 26.39m/s

1) so now 26.39m/s / 6s = 4.4 so acceleration is now 4.4m/s^2

2) d = 1/2 * 4.4 * (6)^2
d = 79.2m
 
To determine the amount of work done by the car, how do I know which formula to use to calculate the force? There is f = m * g and then f = m * a
 
ElegantSir said:
To determine the amount of work done by the car, how do I know which formula to use to calculate the force? There is f = m * g and then f = m * a
The question ought to state this, but I think you can assume it's a horizontal motion.
 
haruspex said:
The question ought to state this, but I think you can assume it's a horizontal motion.
Thank you for the reply. I ended up calculating both then checking which ones were correct online using various work, power, energy calculators to see if they matched the units of emasurement originally given by the equation. Turn outs f = m * a was correct!
 
ElegantSir said:
Thank you for the reply. I ended up calculating both then checking which ones were correct online using various work, power, energy calculators to see if they matched the units of emasurement originally given by the equation. Turn outs f = m * a was correct!
That's not going to help you solve such problems in general. You need to understand why ma is correct here. Did you understand my reply?
 
haruspex said:
That's not going to help you solve such problems in general. You need to understand why ma is correct here. Did you understand my reply?
Yes I understand. Since the car is moving horizontally and not vertically I can assume that the equation should be f = m * a. If the car were motionless then I would need to use f = m * g
 
ElegantSir said:
Yes I understand. Since the car is moving horizontally and not vertically I can assume that the equation should be f = m * a. If the car were motionless then I would need to use f = m * g
No, if the car were motionless then no work would be done, so you would not use either. When would you use g in finding work done?
 
  • #10
haruspex said:
No, if the car were motionless then no work would be done, so you would not use either. When would you use g in finding work done?
Im assuming you would use g only if the car had vertical displacement then?
 
  • #11
ElegantSir said:
Im assuming you would use g only if the car had vertical displacement then?
Yes.
 

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