The Easter Bunny runs along a straight and narrow path with a constant speed of 25 m/s. he passes a sleeping tortoise, which immediately starts to chase the bunny with a constant acceleration of 3 x 10^-3 m/s^2. How long does it take the tortoise to catch the bunny? (Answer in hours)(adsbygoogle = window.adsbygoogle || []).push({});

ok.. so far i've established the following:

Vbunny = 25m/s

a tortoise = 3 x 10^-3

Vi tortoise = 0 m/s

i've got hte formula Vbunny = d/t

i know that i have to use the formula d = Vi t + .5 a t^2

and i know that the distance for both will be the same. thats about as far as it goes.

*new breakthrough*

i have an equation set up:

25m/s t = (0 m/s) t + .5 (3 x 10^-3 m/s^2) t^2

and my next question is basically the same thing:

At the instant when the traffic light turns green, an automobile starts with a constant acceleration of 1.8 m/s^2. At the same time a truck travelling with a constant speed of 8.5 m/s overtakes and passes the automobile. (a) How far beyond the starting point will the automobile overtake the truck? (b) How fast will the car be travelling at that instant?

same question, different values and scenarios. any help would be greatly appreciated

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# Homework Help: More kinematics

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