A car moves with a speed of $60 \mathrm{~km} / \mathrm{hr}$ from point $A$ to point $B$ and then with the speed of $40 \mathrm{~km} / \mathrm{hr}$ from point $B$ to point $C$. Further it moves to point $D$ with a speed equal to its average speed between $A$ and $C$. Points $A, B, C$ and $D$ are…

Question

A car moves with a speed of $60 \mathrm{~km} / \mathrm{hr}$ from point $A$ to point $B$ and then with the speed of $40 \mathrm{~km} / \mathrm{hr}$ from point $B$ to point $C$. Further it moves to point $D$ with a speed equal to its average speed between $A$ and $C$. Points $A, B, C$ and $D$ are collinear and equidistant. The average speed of the car between $A$ and $D$ is

Aicharya · Accepted Answer

Let the points $A, B, C$ and $D$ be separated by $1 \mathrm{~km}$. Then $t_{A B}=\frac{1}{60} h r, t_{B C}=\frac{1}{40} h r$
$\therefore < v_{A C}>=\frac{1+1}{\frac{1}{60}+\frac{1}{40}}=48 \mathrm{~km} / \mathrm{hr} \Rightarrow t_{C D} \frac{1}{48} h r$
$\operatorname{Now} < v_{A D}>=\frac{1+1+1}{\frac{1}{60}+\frac{1}{40}+\frac{1}{48}}=48 \mathrm{~km} / \mathrm{hr}$ *

A car moves with a speed of $60 \mathrm{~km} / \mathrm{hr}$ from point $A$ to point $B$ and then with the…

Solution