Two cyclists, Isabella and Maria, are having a race. They both travel along a straight road with constant acceleration, starting from rest at point $$\(A\)$$. Isabella accelerates for $$\(5 \mathrm{~s}\)$$ at a constant rate $$\(a \mathrm{~m} \mathrm{~s}^{-2}\)$$. She then travels at the constant speed she has reached for $$\(10 \mathrm{~s}\)$$, before decelerating to rest at a constant rate over a period of $$\(5 \mathrm{~s}\)$$. Maria accelerates at a constant rate, reaching a speed of $$\(5 \mathrm{~m} \mathrm{~s}^{-1}\)$$ in a distance of $$\(27.5 \mathrm{~m}\)$$. She then maintains this speed for a period of $$\(10 \mathrm{~s}\)$$, before decelerating to rest at a constant rate over a period of $$\(5 \mathrm{~s}\)$$. Find the value of $$\(a\)$$ for which the two cyclists travel the same distance. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Exam No:9709_s21_qp_41 Year:2021 Question No:4(b)
Answer:
\(\frac{1}{2} a \times 5^{2}+10 \times 5 a+\frac{1}{2} a \times 5^{2}=90\)
or \(\frac{1}{2} \times(20+10) \times 5 a=90\)
\(a=1.2\)
or \(\frac{1}{2} \times(20+10) \times 5 a=90\)
\(a=1.2\)
Knowledge points:
4.2.1 understand the concepts of distance and speed as scalar quantities, and of displacement, velocity and acceleration as vector quantities Restricted to motion in one dimension only. The term ‘deceleration’ may sometimes be used in the context of decreasing speed.
Solution:
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