Minimum Acceleration of an ambulance

  • #1

Homework Statement


An ambulance driver is rushing a patient to the hospital. While traveling at 70 km/h, she notices the traffic light at the upcoming intersections has turned amber. To reach the intersection before the light turns red, she must travel 58 m in 2.5 s.

1.
What minimum acceleration (in m/s2) must the ambulance have to reach the intersection before the light turns red? (Enter the magnitude.)

2.
What is the speed (in km/h) of the ambulance when it reaches the intersection?

Picture of question: https://gyazo.com/cc955179090381952a181d7d7926d3d1



Homework Equations


D = V(initial)t + 1/2at^2[/B]


The Attempt at a Solution


I'm genuinely unsure where to begin. Please, share a couple hints.[/B]
 

Answers and Replies

  • #2
stockzahn
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Homework Equations


D = V(initial)t + 1/2at^2[/B]
This equation already is a very good start to solve question #1. Which values can you assign to the respective symbols in the formula?
 
  • #3
I converted the initial velocity to 19.44 m/s. Then, I inserted the numbers and solved for the acceleration. I derived 3.008 m/s^2 for acceleration. I'm assuming I can insert that to isolate the distance in the original equation?
 
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  • #4
stockzahn
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I converted the initial velocity to 19.44 m/s. Then, I inserted the numbers and solved for the acceleration. I derived 3.008 m/s^2 for acceleration. I'm assuming I can insert that to isolate the final velocity in "vf^2 = vi^2 + 2aD"?
So you've already answered the first question - not a problem obviously. Solving the second question with the work-energy theorem seems a little bit pedestrian (even if it is possible). Maybe you can think about a kinematic correlation to solve the problem, like you did for the first question?
 
  • #5
I calculated 58m for distance and then I inserted that into "vf^2 = vi^2 + 2aD." I then derived 26.96 m/s which is 97.056 km/h. May someone confirm this?
 
  • #6
stockzahn
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Seems correct to me.
 
  • #8
ehild
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I converted the initial velocity to 19.44 m/s. Then, I inserted the numbers and solved for the acceleration. I derived 3.008 m/s^2 for acceleration. I'm assuming I can insert that to isolate the distance in the original equation?
The distance was given, why to isolate it?
How does the velocity change with time ?
 

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