Calculating Time for Dog to Catch Jogger in Grade 11 Physics

[thegreatone09]

Hey! I was wondering if anyone could help me or guide me with this question?

A jogger with a constant velocity of 4.0m/s runs by a stationary dog. After 1.0 seconds, the dog decides to chase the jogger who doesn't realize the dog is chasing ehr. The dog accelerates at constant acceleration of 1.5 m/s/s.
How long does the it take the dog to catch the jogger.

 

[thegreatone09]

please! Anyone?

 

[HallsofIvy]

Do you know the formula for distance for constant acceleration?

You can calculate the position of the runner at any time easily since he is running at constant speed.

You can calculate the position of the dog at any time by using the formula for distance with constant acceleration. (Remember that the dog allows the runner to run for 1 second at 4 m/s before he starts. The dog starts 4 meters behind the runner.)

 

[agro]

Hint 1: If we define the dog to be at d = 0 meters when t = 0 seconds, then at that time the jogger will be at d = 4 meters.

Try to draw diagrams... It'll help you a lot. Another hint is below (try using only the 1st hint)
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Hint 2: When the dog catches the jogger, that means they are at the same position.


Another hint below
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Hint 3: Hint 2 means that at a certain time, the dog and jogger will have the same distance traveled (same d).
Can you solve it? The partial solution is below...
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The equation of motion for the jogger is:
d_jogger = (4 m) + (4 m/s) * t

For example at time t = 1 we have
d_jogger = (4 m) + (4 m/s) * (1 s)
= 8 m

The equation of motion for the dog is:
d_dog = 0.5 * (1.5 m/s/s) * t²

For example at time t = 1 we have
d_dog = 0.5 * (1.5 m/s/s) * (1 s)²
= 0.75 m

Obviously at time t = 1 s, the dog haven't catched the jogger. We can go on and try t = 2 s, t = 3 s, t = 4 s, etc, but that would be ineffective since we can use a faster method... We can try to derive an equation using the known facts, and solve it.

From hint 3, we know that at a certain time (the time when the dog catches the jogger), the dog and jogger will have the same distance traveled (same d). That means d_jogger = d_dog

So we write

d_jogger = d_dog
(4 m) + (4 m/s) * t = 0.5 * (1.5 m/s/s) * t²

Now you only need to solve for t...

Does any part dazzle you? For example do you know how I "magically" create those 2 equation of motions? Do you know how to solve the final equation? If you solve it, you'll get 2 solutions (two different time when the dog catches the guy!). Weird? Do you know why there are 2 solutions?

Have fun!