Question

A car is moving on a straight road from $A$ to $B$ for first one-fourth distance with speed $40m/s$ and the next half with speed $80m/s$ and the last one-fourth with speed $120m/s$. Then, the average speed of the car will be

• A. $49.26m/s$
• B. $90.46m/s$
• C. $68.57m/s$
• D. $54.26m/s$

Answer 1

Answer: (C)

According to the question, the situation is as shown,

where, $d=$ total distance between $A$ and $B$.
From $A$ to $C$,
Time taken, $t_1=\frac{d/4}{40}=\frac{d}{160}$
From $C$ to $D$,
Time taken, $t_2=\frac{d/2}{80}=\frac{d}{160}$
From $D$ to $B$,
Time taken, $t_3=\frac{d/4}{120}=\frac{d}{480}$
Total time $=t_1+t_2+t_3=\frac{d}{160}+\frac{d}{160}+\frac{d}{480}$
$=\frac{3d+3d+d}{480}=\frac{7d}{480}$
Average speed $=\frac{\text{ Total distance }}{\text{ Total time }}=\frac{d}{7d/480}$
$=\frac{480}{7}=68.57m/s$