# Kinematics in one dimension (easy problem)

Im new to this and i know this must be a very simple problem. I cant seem to find any formulas for distance or a way of doing this problem. Most of the help sites give you the distance and want you to find something else. Could someone help me understand what i need to do to solve this. Thank you very much. Any tips, sites, or anything you can think of that would help me get through my courses and help me understand would be greatly appriciated.

A tourist being chased by an angry bear is running in a straight line towards his car at a speed of 4.0ms. The car is a distance of d away. The bear is 26m behind the tourist and running at 6.0 ms. The tourist reaches the car safely. What is the maximum possible value for d?

Well for the tourist to reach his car safely then

$$4.0t$$

Has to be greater than

$$6.0t - 26$$

Do you see why? Some algebra 1 will give you the time he can run before the bear catches up. Then use this time and the runner's speedto find out the distance he can run before he gets caught.

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I do not believe there is anything wrong with what whozum has said but I would like to put another option forward.

The driver must run $$4.0 \ ms^{-1} \times t = d \ metres$$.

The bear must run $$6.0 \ ms^{-1} \times t = (d + 26) metres$$ to catch up.

Substitute the first equation into the second and work out t. This will give the time that the man must run before being caught up. Then you can work out a distance that the man has left to run before being caught and that will be less than the safe distance (if you consider he has to get in the car and start the engine ).

Thank you very much. Just wondering is that a formula i can use for finding out distance im most cases?

d = v t for constant velocity

d = v t + .5 at^2 for constant acceleration

Or using calculus you can derive it all from

$$\int_{t_0}^{t_1} a\ dt = \frac{dx}{dt}$$

$$\int_{t_0}^{t_1} a\ dt = \frac{dx}{dt}$$ which is equal to the distance travelled???