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Question

Question: The temperature dropped 18 degrees Celsius in the last 24 days. If the rate of temperature drop rema...

The temperature dropped 18 degrees Celsius in the last 24 days. If the rate of temperature drop remains the same, how many degrees will the temperature drop in the next 18 days?

Explanation

Solution

In the question, we need to determine the total drop in the temperature after 18 days from the 24 days (reference). For this, we first need to determine the rate of change of temperature per day and then multiply with the total number of days to get the result.

Complete step by step solution:
To determine the rate of change of temperature per day, divide the change in temperature to the change in days.
So, substitute δθ=18\delta \theta = - 18 (negative sign is used as the value of the temperature is reducing) and δt=24\delta t = 24 in the equation R=δθδtR = \dfrac{{\delta \theta }}{{\delta t}} to get the rate of change in temperature.
R=δθδt =1824 =34(i) R = \dfrac{{\delta \theta }}{{\delta t}} \\\ = \dfrac{{ - 18}}{{24}} \\\ = \dfrac{{ - 3}}{4} - - - - (i)
This is the change in temperature per day.
Now, according to the question, we need to determine the temperature after 18 days of 24 days, i.e., δt=24+18=42\delta t' = 24 + 18 = 42.
To get the temperature drop after 42 days starting from 0th day, multiply the rate of change in temperature with the total number of change in days as:
δθ=δθδt×δt =34×42 =31.5 \delta \theta ' = \dfrac{{\delta \theta }}{{\delta t}} \times \delta t' \\\ = \dfrac{{ - 3}}{4} \times 42 \\\ = - 31.5
\therefore temperature drop in 42 days = -31.5 degrees.

**Hence, the temperature change in the last 18 days is -31.5-(-18) = -13.5 degrees.

Note: **
It should be noted down here that the temperature drop has been given in the question so, the rate of change of the temperature will come into negation. In other words, we can say that as the initial value of the temperature is more than the final value; hence, the rate of change of temperature per day comes out to be negative.