A particle moves towards east from a point $A$ to a point $B$ at the rate of $4 \mathrm{~km} / \mathrm{h}$ and then towards north from $B$ to $C$ at the rate of $5 \mathrm{~km} / \mathrm{h}$. If $A B=12 \mathrm{~km}$ and $B C=5 \mathrm{~km}$, then its average speed for its journey from $A$ to $C$ and resultant average velocity direct from $A$ to $C$ are respectively

(A) $\frac{17}{4} \mathrm{~km} / \mathrm{h}$ and $\frac{13}{4} \mathrm{~km} / \mathrm{h}$

(B) $\frac{13}{4} \mathrm{~km} / \mathrm{h}$ and $\frac{17}{4} \mathrm{~km} / \mathrm{h}$

(C) $\frac{17}{9} \mathrm{~km} / \mathrm{h}$ and $\frac{13}{9} \mathrm{~km} / \mathrm{h}$

(D) $\frac{13}{9} \mathrm{~km} / \mathrm{h}$ and $\frac{17}{9} \mathrm{~km} / \mathrm{h}$

JEE Main · 2004 · 24 Apr 2004 · Offline · Chapter_pending

A particle moves towards east from a point

A

to a point

B

at the rate of

4 km / h

and then towards north from

B

C

at the rate of

5 km / h

. If

A B = 12 km

and

B C = 5 km

, then its average speed for its journey from

A

C

and resultant average velocity direct from

A

C

are respectively

Answer: (A)

Going east, the time for

A B

\frac{12}{4} = 3

hours; going north, the time for

B C

\frac{5}{5} = 1

hour, so the whole trip takes

4

hours. The total path length is

12 + 5 = 17

km, giving an average speed of

\frac{17}{4}

km/h. The straight-line displacement is

A C = 1 2^{2} + 5^{2} = 13

km, so the average velocity from

A

C

has magnitude

\frac{13}{4}

km/h.

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option