Set 2 Problem number 3

The velocity of an object increases by 8 meters per second during its lifetime.

If the rate of increase is constant, and if during a 1 second lifetime another object with the same velocity is observed to increase its velocity be 25 meters / sec, then how long does this object live?

Solution

Every second the velocity increases by 25 meters per second.

The increase here is less that 25 meters/second, so it will take less than a second to accomplish this increase.
Simply thinking about the ratios tells us that it will take only 8 / 25 second to increase by 8 meters per second.
The answer is thus ( 8 meters/sec) / ( 25 meters/sec/sec) = .32 sec.

This reasoning is equivalent to the idea of a rate, where we find the time required by dividing the change in the quantity by the rate.

In this case the changing quantity is velocity and the rate is the acceleration, so we have

time required = change in quantity / rate = `dt = `dv / a.

Given the increase of velocity per second, i.e. the rate of velocity increase which is the acceleration a , and the increase `dv in velocity, the increase `dv must be equal to the product a `dt of the per second increase and the number of seconds `dt .

Since `dv = a `dt, we have `dt = a / `dv.

Explanation in terms of Figure(s), Extension

Figure description:

We can again visualize the aAve, `dt, `dv relationships as a triangle. We would in this case use `dt = `dv / aAve.