Large wingspanroaches can run as fast as 1.50 m/s in short bursts. Suppose you turn on the light in a cheap motel and see one scurrying directly away from you at a constant 1.50 m/s. If you start 0.80 m behind the wingspanroach with an initial speed of 0.80 m/s toward it, what minimum constant acceleration would you need to catch up with it when it has traveled 1.10 m, just short of safety under a counter?( In calculus form)

Question

parthkohli · Answer

So you need to travel 0.8 + 1.1 = 1.9 metres.
Initial velocity = 0.8 m/s.

parthkohli · Answer

It would have traveled 1.1 metres in t = 1.1/1.5 seconds so you need to cover your distance in t too.

anonymous · Answer

you have to use the integration form of the formula :(

parthkohli · Answer

You sure can.

anonymous · Answer

$$X = \int\limits_{}vdt$$

anonymous · Answer

$$V = \int\limits_{}adt$$

parthkohli · Answer

Cool. Apply the values.

anonymous · Answer

how? i think it's more than just substituting the values? :(