Question Video: Evaluating Absolute Value Functions by Direct Substitution | Nagwa Question Video: Evaluating Absolute Value Functions by Direct Substitution | Nagwa

Question Video: Evaluating Absolute Value Functions by Direct Substitution Mathematics • Second Year of Secondary School

A body was moving with a uniform velocity of magnitude 5 cm/s from the point 𝐴 to the point 𝐶 passing through the point 𝐵 without stopping. The distance between the body and point 𝐵 is given by 𝑑(𝑡) = 5|8 − 𝑡|, where 𝑡 is the time in seconds and 𝑑 is the distance in cm. Determine the distance between the body and the point 𝐵 after 5 seconds and after 11 seconds.

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Video Transcript

A body was moving with a uniform velocity of magnitude five centimeters per second from the point 𝐴 to the point 𝐶 passing through the point 𝐵 without stopping. The distance between the body and point 𝐵 is given by 𝑑 of 𝑡 is equal to five multiplied by the absolute value of eight minus 𝑡, where 𝑡 is the time in seconds and 𝑑 is the distance in centimeters. Determine the distance between the body and the point 𝐵 after five seconds and after 11 seconds.

We are given a diagram which shows the body that is about to move from point 𝐴 to point 𝐶 via point 𝐵 with a velocity of five centimeters per second. Whilst there is a lot of information in this question, the key point is that the function 𝑑 of 𝑡 is equal to five multiplied by eight minus 𝑡. 𝑑 of 𝑡 is the distance of the body from point 𝐵 after a given time. We need to calculate this distance after five seconds and also after 11 seconds. After five seconds, 𝑡 is equal to five. Therefore, we need to calculate 𝑑 of five.

This is equal to five multiplied by the absolute value of eight minus five. Eight minus five is equal to three, so we need to multiply five by the absolute value of three. As the absolute value of a number is its distance from zero, the absolute value of three is three. As five multiplied by three is equal to 15, the distance between the body and the point 𝐵 after five seconds is 15 centimeters.

We need to repeat this process when 𝑡 equals 11. This means we need to calculate the value of 𝑑 of 11. This is equal to five multiplied by the absolute value of eight minus 11. Eight minus 11 is equal to negative three. As the absolute value of a number is its distance from zero, the absolute value of negative three is also three. In fact, the absolute value of any number will always be positive. Multiplying five by three once again, we see that the distance between the body and point 𝐵 after 11 seconds is also 15 centimeters.

In terms of our diagram, we can see that after five seconds and 11 seconds, the body is the same distance away from point 𝐵. After five seconds, it is still approaching point 𝐵 from point 𝐴. And after 11 seconds, it is past point 𝐵 and it’s heading towards Point 𝐶.

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