What is Maddie's final velocity after accelerating for 10 seconds?

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swede5670
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Homework Statement


Flash Maddie is running at a whopping velocity of 30 mph. She then sees Mr. T driving down the road straight towards her so she accellerates -5 mph/s for 10 seconds. what Is Maddie's velocity now. (In mph)

Homework Equations


V = Change in distance / change in time
a = change in velocity / change in time


The Attempt at a Solution


I'm just not sure how to approach the problem in general and I'm not sure how to approach the negative acceleration.
 
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swede5670 said:

Homework Statement


Flash Maddie is running at a whopping velocity of 30 mph. She then sees Mr. T driving down the road straight towards her so she accellerates -5 mph/s for 10 seconds. what Is Maddie's velocity now. (In mph)

Homework Equations


V = Change in distance / change in time
a = change in velocity / change in time


The Attempt at a Solution


I'm just not sure how to approach the problem in general and I'm not sure how to approach the negative acceleration.
Whilst you could in principle use those equations, this question is best solved using kinematics equations (of uniform acceleration).
 
Use this equation:
a = change in velocity / change in time

Plug in the numbers you know, including any negative signs, and solve the equation. That will give you a key piece of information for solving the problem.
 
Hootenanny: What are these equations and how would I use them?
Red Belly: Should I multiply -5 MPH/s by 10 and then subtract it from 30 mph?
 
Last edited:
swede5670 said:
Hootenanny: What are these equations and how would I use them?
Red Belly: Should I multiply -5 MPH/s by 10 and then subtract it from 30 mph?
Yes, that's correct (with one minor correction, you should add the -50 to 30). In actual fact, both methods are identical:

[tex]a=\frac{\Delta v}{\Delta t} = \frac{v_f-v_i}{\Delta t}[/tex]

[tex]\Rightarrow v_f = v_i + a\Delta t[/tex]

Which is one of the kinematic equations I was referring to.
 
It may be that swede's class has not quite gotten to the full set of kinematic equations ... at any rate, Hootenanny is entirely right.