Driver sees stop sign up ahead

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David is driving at 25 m/s and sees a stop sign 10 meters ahead, applying brakes with a deceleration of 4 m/s². The key question is whether he can stop within the 10-meter distance before reaching the sign. A kinematic equation that does not involve time can be used to determine if he stops in time. The discussion clarifies that the time to stop is irrelevant; the focus is solely on the stopping distance. Ultimately, the problem was resolved, confirming that David can stop before reaching the sign.
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Homework Statement



I am having some trouble with a question. Here it is:

David is driving along at 25m/s when he sees a stop sign ahead. He hits the breaks at 10m from the sign. He decelerates at 4m/s-squared. Does he stop in time?

Homework Equations



We were given the equations of motion. The question gives displacement, initial velocity, and acceleration.

The Attempt at a Solution



My main gripe with this question is that it doesn't give a time that he must reach. Is another car going to reach the traffic stop in 10 seconds? 15 seconds? It doesn't say...

I thought of simply seeing how long it would take him to reach it if instead of braking he kept going at constant velocity but no luck.

Can someone help me out?
 
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Ignore the time -- in this case it doesn't matter. The question only asks if he can completely stop in 10m. There just so happens to be a kinematic equation that doesn't contain time. Finding that equation makes this a (relatively) simple algebra problem.
 
time is not asked. by 'does he stop in time' it is meant whether he stops within 10m distance.
 
You could calculate the time required to stop from the formula [math]v= -at+ v_0[/math], taking v= 0. Then use the distance formula x= -(a/2)t^2+ v_0t[/math] to determine whether he &quot;stopped in time&quot;- whether he stopped before going 10 m.<br /> <br /> By the way, he hit the &quot;brakes&quot;, not the &quot;breaks&quot;!
 
HallsofIvy said:
You could calculate the time required to stop from the formula [math]v= -at+ v_0[/math], taking v= 0. Then use the distance formula x= -(a/2)t^2+ v_0t[/math] to determine whether he &quot;stopped in time&quot;- whether he stopped before going 10 m.<br /> <br /> By the way, he hit the &quot;brakes&quot;, not the &quot;breaks&quot;!
<br /> <br /> Hehehe yeah. I was in a hurry. &gt;.&lt; That&#039;s an embarrassing mistake. <br /> <br /> Thanks for the help everyone. Figured it out... don&#039;t how I didn&#039;t realize that right away.
 
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