Question

A particle goes from $A$ to $B$ with a speed of $40km{h}^{-1}$ and $B$ to $C$ with a speed of $60km{h}^{-1}$ . If $AB=6BC$ , the average speed in $km{h}^{-1}$ between $A$ and $C$ is

Answer 1

Answer: 42

$V_{av}=\frac{\text{distance travelled}}{\text{time taken}}$
$\Rightarrow V_{av}=\frac{AB+BC}{\frac{AB}{40}+\frac{BC}{60}}$
$\Rightarrow V_{av}=\frac{6BC+BC}{\frac{6BC}{40}+\frac{BC}{60}}$
$V_{av}=\frac{7\times 240}{36+4}\Rightarrow V_{av}=\frac{7\times 240}{40}$
$\Rightarrow V_{av}=42km/h$