Solving for Car's Acceleration, Time, and Average Speed

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SUMMARY

The discussion focuses on calculating a car's acceleration, time, and average speed given an initial rest state, a distance of 1250 meters, and a final speed of 112 km/h. The user initially calculated the acceleration as 0.77 m/s² using the formula A = delta V / delta T, resulting in a time of 40.18 seconds. However, a second calculation using the equation v² = V(initial)² + 2a(d - d(initial)) yielded an acceleration of 0.387 m/s², which raised concerns about the discrepancy. The correct approach emphasizes that the car accelerates over the entire distance before reaching its final speed.

PREREQUISITES
  • Understanding of kinematic equations, specifically v² = V(initial)² + 2a(d - d(initial))
  • Knowledge of basic physics concepts related to acceleration and motion
  • Familiarity with unit conversions, particularly converting km/h to m/s
  • Ability to manipulate algebraic equations to solve for unknowns
NEXT STEPS
  • Review the kinematic equations of motion and their applications in physics
  • Learn how to convert speeds from km/h to m/s accurately
  • Practice solving problems involving acceleration and distance using different kinematic equations
  • Explore the concept of average speed and its calculation in uniformly accelerated motion
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of kinematic problem-solving techniques.

Goopy17
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Homework Statement


A car is initial at rest. It accelerates over a distance of 1250m and reaches a speed of 112 km/h. Find: A) the acceleration for the car B) the time it took for it to reach this final speed C) the average speed


Homework Equations


A = delta V/ delta T or v^2 = V (initial)^2 + 2a (d -d(initial))
d = d(initial) + vt



The Attempt at a Solution


1250 = 0 + 31.1(t)
t = 40.18

A = 31.1/40.18
A = .77 m/s^2

BUT here is the problem

if i use v^2 = V (initial)^2 + 2a (d -d(initial))
31.1^2 = 0^2 + 2a(1250)
A = .387 m/s^2
that is half of the first one?


that is where i need help
Thanks!
 
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Goopy17 said:
if i use v^2 = V (initial)^2 + 2a (d -d(initial))
31.1^2 = 0^2 + 2a(1250)
A = .387 m/s^2
that is half of the first one?

Your solution for the acceleration in this section is correct.

Goopy17 said:
1250 = 0 + 31.1(t)
t = 40.18

A = 31.1/40.18
A = .77 m/s^2

No.
Remember, the car is accelerating over this distance. It travels the 1250 meters before it reaches it's final speed.

Do you have the other equations of motion there? Which of them can you use that uses time and acceleration to compute a distance?
 
Not sure. Can you please explain?
 

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