Question

Two boats A and B are moving along perpendicular paths in a still lake at night. Boat A move
a speed of 3m/s and boat B moves with a speed of 4 m/s in the direction such that they collide
after sometime. At t = 0, the boats are 300 m apart. The ratio of distance travelled by boat A to
the distance travelled by boat B at the instant of collision is:
(A) 1
(B) 1/2
(C) 3/4
(D) 4/3
1
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Answer 1

Answer:

Ratio of the distance moved by the two boats is given as

$d_1 : d_2 = 3 : 4$

Explanation:

As we know that two boats move in perpendicular directions and then collide

So here the distance moved by the boats is given as

distance = (speed)(time)

for boat moving with speed 3 m/s

$d_1 = 3 t$

for boat moving with speed 4 m/s

$d_2 = 4 t$

now we know that the ratio of of the distance moved by two boats is given as

$d_1 : d_2 = 3t : 4t$

$d_1 : d_2 = 3 : 4$

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Topic : Kinematics

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Answer 2

Answer:

Ratio of the distance moved by the two boats is given as

d_1 : d_2 = 3 : 4d

1

:d

2

=3:4

Explanation:

As we know that two boats move in perpendicular directions and then collide

So here the distance moved by the boats is given as

distance = (speed)(time)

for boat moving with speed 3 m/s

d_1 = 3 td

1

=3t

for boat moving with speed 4 m/s

d_2 = 4 td

2

=4t

now we know that the ratio of of the distance moved by two boats is given as

d_1 : d_2 = 3t : 4td

1

:d

2

=3t:4t

d_1 : d_2 = 3 : 4d

1

:d

2

=3:4

Explanation:

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