Question

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

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

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 $=t_{1}+t_{2}+t_{3}=\frac{d}{160}+\frac{d}{160}+\frac{d}{480}$
$=\frac{3 d+3 d+d}{480}=\frac{7 d}{480}$
Average speed $=\frac{\text { Total distance }}{\text { Total time }}=\frac{d}{7 d / 480}$
$=\frac{480}{7}=68.57 \,m / s$