Question

During the first $18$ min of a $60$ min trip, a car has an average speed of $11m{s}^{-1}.$ What should be the average speed for remaining $42$ min so that car is having an average speed of $21m{s}^{-1}$ for the entire trip?

• A. $25.3m{s}^{-1}$
• B. $29.2m{s}^{-1}$
• C. $31m{s}^{-1}$
• D. $35.6m{s}^{-1}$

Answer 1

The correct option is A $25.3m{s}^{-1}$

Distance covered in first 18 min = $11\times 18\times 60=11880m$
Let average speed for remaining trip be 'v'.
Thus distance covered in remaining time = $v\times 42\times 60=2520vm$
Thus total distance, $D=11880+2520v$
total time, $T=60min=3600sec$
Thus $Avg.Speed=21=\frac{D}{T}(11880+2520v)=21\times 3600v\approx 25.3m/s$