Calculate velocity of stopping car

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SUMMARY

The discussion centers on calculating the velocity of a car that decelerates from an initial speed of 20 m/s to a stop over a distance d. Participants emphasize using the kinematic equation \( v^2 = u^2 + 2as \) to find the velocity at distance \( d/2 \). The acceleration (a) is not directly provided, but can be derived from the problem's parameters. The final solution involves applying the formula \( v = \sqrt{u^2 + 2as} \) to determine the car's velocity at the midpoint of its stopping distance.

PREREQUISITES
  • Understanding of kinematic equations in physics
  • Familiarity with concepts of acceleration and deceleration
  • Basic algebra for manipulating equations
  • Knowledge of initial and final velocity definitions
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  • Study the kinematic equation \( v^2 = u^2 + 2as \) in detail
  • Learn how to derive acceleration from distance and velocity
  • Explore real-world applications of deceleration in automotive physics
  • Practice solving problems involving motion under constant acceleration
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Students studying physics, particularly those focusing on kinematics, as well as educators looking for problem-solving techniques in motion analysis.

Lizi
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Homework Statement


A car moving at 20m/s starts decelerating, travels distance d, and stops. Find the car’s velocity at distance d/2.

Homework Equations

The Attempt at a Solution


my brain is fried. I feel like I’m missing something obvious but I just don’t get it.
 
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I suppose you should assume a constant deceleration
 
Lizi said:

Homework Statement


A car moving at 20m/s starts decelerating, travels distance d, and stops. Find the car’s velocity at distance d/2.

Homework Equations

The Attempt at a Solution


my brain is fried. I feel like I’m missing something obvious but I just don’t get it.

It was a bit of a brain 'fry' !

Time 'reversing' so that we have acceleration from 0 m/s instead of deceleration from 20 m/s:

$$ d = ½at^2 ⇒ \frac{d}{2} = ½a\left({\frac{t}{\sqrt{2}}}\right)^2 ⇒v=\frac{at}{\sqrt{2}}$$
 
neilparker62 said:
It was a bit of a brain 'fry' !

Time 'reversing' so that we have acceleration from 0 m/s instead of deceleration from 20 m/s:

$$ d = ½at^2 ⇒ \frac{d}{2} = ½a\left({\frac{t}{\sqrt{2}}}\right)^2 ⇒v=\frac{at}{\sqrt{2}}$$

But since you don't know ##a## you can't obtain ##t##, so you stick with two unknowns. I think the question should be solved independent of the time ...
 
Edit: This cross posted with some of the above;

It gives you the initial velocity (u) and asks you to find the cars velocity (v) after traveling a distance (s = d/2). What equation relates u,v,s and a?

You aren't given the acceleration (a) but you can work it out from the first part of the problem statement.
 
stockzahn said:
But since you don't know ##a## you can't obtain ##t##, so you stick with two unknowns. I think the question should be solved independent of the time ...
at = ?
 
neilparker62 said:
at = ?

Still, I think the medthod @CWatters proposes is the preferred one. Additionally it should be the OP's task to solve it ...
 
guys, I figured it out using the d=(v2-u2)/(2a) formula.
Thank you all so much
 
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