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I have tried my best to understand why I get different answers for the following questions. Why does the equation matter?

Question 1: A groundhog running at $2.00\space ms^{-1}$ passed a dog which proceeded to chase it. How long did it take for the dog to catch up if it accelerated at $0.450 \space ms^{-2}$ the whole way?

Answer: $8.89s$

Method 1:

$v_1 = 0$

$v_2 = 2.00m/s$

$a = 0.450ms^{-2}$

$t = ?$

$t = v_2 - \frac{v_1}{a} = 4.44s$

Method 2:

$d = vt$

$vt = v_1t + \frac{1}{2}at^2$

$t = \frac{2v}{a} = 8.89s$ $$$$

  1. Why is it that different equations produce different results?

  2. I'm really frustrated by this issue, am I doing something wrong?

  3. If possible, can someone tell me when to actually use each kinematic equation and in which situation?

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    $\begingroup$ Could you edit this to add in what you think each of the variables in the equations means? It's very easy to get confused when you jump straight to the equations rather than worrying about what they mean. For example, "Way 1" is likely focused on how long it takes to reach 2.0m/s, which is the same speed as the groundhog, but Woof is already behind, just running in parallel, so he needs to keep accelerating even further to catch up. I think if you add more words, you'll find the answer is already there. You're quite close! $\endgroup$
    – Cort Ammon
    Commented Jan 24, 2017 at 2:03
  • $\begingroup$ - Cort Ammon, your words were so pure they made me realize the mistake, so really way 1 is just the how long it takes the hog to get 2.00m/s when reaching Woof which is obviously half of the full distance travelled . In the second way distance is substituted since it doubles and you have to find it for Woof not for the hog . Hopefully this thinking is right. Thanks $\endgroup$
    – lordzee
    Commented Jan 24, 2017 at 2:16

3 Answers 3

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In method 1 you found the time taken for the dog to reach the speed of the ground hog. During that time the ground hog has traveled twice the distance traveled by the dog as is shown in the velocity time graph below.

enter image description here

Area under graph = distance covered.

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I really don't know how you got your first greyish equation but it makes no sense dimensionally: on the left hand side you have units of time and the first term in the right has units of length/time. I suspect that you meant $t=(v_2-v_1)/a$, which would apparently give you an answer compatible with your second calculation.

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  • $\begingroup$ It is the same equation you mentioned and it does not provide the same answer as the second one. $\endgroup$
    – lordzee
    Commented Jan 24, 2017 at 2:18
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Actually there is a different approach to this question.
IN way 1,the time taken for dog to accelerate is 4.44 seconds.Cacullate the total distance travelled by dog during that time and also calculate the total distance travelled by hog at that time.Now given that they both are at equal velocities,now calculate the time taken for dog to catch the hog.Now add this time with the time you got before.

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