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Case Study - 1
Two motorboats $A$ and $B$ are waiting at the opposite banks of a river in order to reach the opposite side. From a point $P$ on the bridge, $20\text{ m}$ above the river, the angles of depression of the boats are $30^\circ$ and $45^\circ$ respectively, as shown in the figure given below. Both the boats leave at the same time at the speed of $10\text{ m/s}$ and $5\text{ m/s}$, respectively.
Based on the above information, answer the following questions :
(i) Find the distance travelled by boat $A$ to reach point $D$ in the river, vertically below the point $P$. (Use $\sqrt{3} = 1.73$)
(ii) What is the width of the river?
(iii) (a) Which boat will reach point $D$ first, and how much earlier, than the other boat?
OR
(iii) (b) What is the distance between the two boats after $3$ seconds?
Two motorboats $A$ and $B$ are waiting at the opposite banks of a river in order to reach the opposite side. From a point $P$ on the bridge, $20\text{ m}$ above the river, the angles of depression of the boats are $30^\circ$ and $45^\circ$ respectively, as shown in the figure given below. Both the boats leave at the same time at the speed of $10\text{ m/s}$ and $5\text{ m/s}$, respectively.
Based on the above information, answer the following questions :
(i) Find the distance travelled by boat $A$ to reach point $D$ in the river, vertically below the point $P$. (Use $\sqrt{3} = 1.73$)
(ii) What is the width of the river?
(iii) (a) Which boat will reach point $D$ first, and how much earlier, than the other boat?
OR
(iii) (b) What is the distance between the two boats after $3$ seconds?
Show SolutionHide Solution↓
(i) $\frac{20}{AD} = \tan 30^\circ \Rightarrow AD = 20\sqrt{3} = 34.6\text{ m}$ [$\frac{1}{2} + \frac{1}{2}$ mark]
(ii) $\frac{20}{DB} = \tan 45^\circ \Rightarrow DB = 20\text{ m}$ [$\frac{1}{2}$ mark]
$\text{Width of river} = 34.6 + 20 = 54.6\text{ m}$ [$\frac{1}{2}$ mark]
(iii) (a) $\text{Time taken by boat } A = \frac{34.6}{10} = 3.46\text{ seconds}$ [$1$ mark]
$\text{Time taken by boat } B = \frac{20}{5} = 4\text{ seconds}$ [$\frac{1}{2}$ mark]
$\text{Boat } A \text{ will reach earlier by } 0.54\text{ seconds}$ [$\frac{1}{2}$ mark]
OR
(iii) (b) $\text{Distance covered by boat } A \text{ in } 3 \text{ seconds} = 3 \times 10 = 30\text{ m}$ [$\frac{1}{2}$ mark]
$\text{Distance covered by boat } B \text{ in } 3 \text{ seconds} = 3 \times 5 = 15\text{ m}$ [$\frac{1}{2}$ mark]
$\text{Distance between them after } 3 \text{ seconds} = 54.6 - (30 + 15) = 9.6\text{ m}$ [$1$ mark]
(ii) $\frac{20}{DB} = \tan 45^\circ \Rightarrow DB = 20\text{ m}$ [$\frac{1}{2}$ mark]
$\text{Width of river} = 34.6 + 20 = 54.6\text{ m}$ [$\frac{1}{2}$ mark]
(iii) (a) $\text{Time taken by boat } A = \frac{34.6}{10} = 3.46\text{ seconds}$ [$1$ mark]
$\text{Time taken by boat } B = \frac{20}{5} = 4\text{ seconds}$ [$\frac{1}{2}$ mark]
$\text{Boat } A \text{ will reach earlier by } 0.54\text{ seconds}$ [$\frac{1}{2}$ mark]
OR
(iii) (b) $\text{Distance covered by boat } A \text{ in } 3 \text{ seconds} = 3 \times 10 = 30\text{ m}$ [$\frac{1}{2}$ mark]
$\text{Distance covered by boat } B \text{ in } 3 \text{ seconds} = 3 \times 5 = 15\text{ m}$ [$\frac{1}{2}$ mark]
$\text{Distance between them after } 3 \text{ seconds} = 54.6 - (30 + 15) = 9.6\text{ m}$ [$1$ mark]