2022 EJC J1 MYE Q10

2022 EJC J1 MYE Q10

10 marks
Mid Year Exam

In a particular video animation, an animal \(X\) chases other animals. All animals only move in leaps in a straight line. You may assume that the time taken for each leap is the same and that the time in between leaps is negligible.

Animal \(X\) starts is chase by taking the first leap with a distance of \(\mathrm{5}\)m and each of its subsequent leaps is \(\mathrm{0}\mathrm{.02}\)m less than the previous.

  1. Find the distance covered in the \(\mathrm{20}\)th leap and find the total distance covered in the first \(\mathrm{20}\) leaps by animal \(X\).[3]

Animal \(Y\) escapes from animal \(X\) by taking its first leap with a distance of \(\mathrm{5}\)m and each of its subsequent leaps is \(2\% \) shorter than the previous.

  1. Given that animal \(Y\) is \(20\)m away from animal \(X\) at the start of the chase, find the least number of leaps for animal \(X\) to overtake animal \(Y\).[4]

Another animal \(Z\) escapes from animal \(X\) by taking leaps of constant distance \(\mathrm{4}\)m.

  1. Given that animal \(Z\) is \(\mathrm{30}\)m away from animal \(X\) at the start of the chase, show that animal \(X\) will never catch animal \(Z\).[3]

Video Solution:

Video Solution

Video solution locked

A free account saves your progress; this video needs tuition enrolment.

Solution:

Solution locked

680+ questions here come with free worked solutions. A free account saves your questions and progress; this one needs tuition enrolment.

Create a free account to practise similar questions.

Answer:(a) T₂₀ = 4.62 m, S₂₀ = 96.2 m (b) 26 leaps (c) Animal X will never catch animal Z

Need help? Join our JC Math tuition classes.

Learn more