Calculating Distance and Velocity: Solving Kinematics Problems

[bobby_burke]

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)

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. that's 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 traveling 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 traveling at that instant?

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

 

[jamesrc]

For the first one, just solve the equation you wrote for t and that is the answer.

For the second one, set it up the same way:
$$\frac{1.8}{2}t^2=8.5t$$

Once you've solved for t, you can use that to find the distance traveled (8.5*t from the constant speed of the truck) and for the car's speed (v = v_o+at)

ETA: Thanks tony873004

 

[tony873004]

Just fixing jamesrc's tex code :)
$$\frac{1.8}{2}t^2=8.5t$$

 

[bobby_burke]

for time i broke the equation down into:
8.5 m/s = t
.5(1.8m/s^2)

and my time was 9.4 seconds... sound right?

 

[bobby_burke]

thanks guys, you have been a big help tonight

 

[tony873004]

That sounds right for t. But the question asks for distance and velocity. Did you get them yet?