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 to 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

Correct answer: A. (17)/(4) ~km / h and (13)/(4) ~km / h Time T_1 from A to B=(12)/(4)=3 hrs T_2 from B to C=(5)/(5)=1 hrs . Total time =4 hrs . Average speed =(17)/(4) ~km / hr . Resultant average velocity =(13)/(4) ~km / hr .

JEE Main 2004

Exam: JEE Main
Year: 2004
Subject: Mathematics
Chapter: Heights and Distances

JEE Main 2004 · Mathematics · Heights and Distances

A particle moves towards east from a point A to a point B at the...

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

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

C

and resultant average velocity direct from

A

C

are respectively 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 to C

Options

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 $\frac{17}{9} \mathrm{~km} / \mathrm{h}$

Correct answer

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

Step-by-step solution

Time

T_1

from

A

B=\frac{12}{4}=3 \mathrm{hrs}

T_2

from

B

C=\frac{5}{5}=1 \mathrm{hrs}

. Total time

=4 \mathrm{hrs}

. Average speed

=\frac{17}{4} \mathrm{~km} / \mathrm{hr}

. Resultant average velocity

=\frac{13}{4} \mathrm{~km} / \mathrm{hr}

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