Question

Two particles A and B are at rest at separation
of 160 m. Both the particle starts moving at t =
0.Particle A moves in a straight line
perpendicular to initial position AB with constant
acceleration of 3 m/s2. Particle B moves in
such a way that it always aims towards A and
moves with speed increasing at constant rate of
5 m/s2. Then
A.
B
(1) The time elapsed when both particles meet
is 5 s
(2) The time elapsed when both particles meet
is 10 s
(3) Distance covered by A before meeting is 150 m
(4) Distance covered by A before meeting is 200 m​

Answer 1

Given that,

Distance = 160 m

Acceleration of A = 3 m/s²

Acceleration of B = 5 m/s²

Suppose, The time elapsed when both particles meet is and distance covered by a before meeting

We need to calculate the distance of A

Using equation of motion

$s_{A}=ut+\dfrac{1}{2}at^2$

$s_{A}=0+\dfrac{1}{2}\times3\times t^2$.....(I)

We need to calculate the distance of B

Using equation of motion

$s_{B}=ut+\dfrac{1}{2}at^2$

$s_{B}=0+\dfrac{1}{2}\times5\times t^2$....(II)

We need to calculate the total distance

Using formula for total distance

$s_{B}^2=160^2+s_{A}^2$

Put the value into the formula

$(\dfrac{1}{2}\times5\times t^2)^2=160^2+(\dfrac{1}{2}\times3\times t^2)^2$

$\dfrac{25}{4}t^4-\dfrac{9}{4}t^4=160^2$

$\dfrac{16}{4}t^4=160^2$

$t^2=\dfrac{160\times2}{4}$

$t=\sqrt{80}$

$t=8.9\ sec$

We need to calculate the distance covered by A before meeting

Using equation (I)

$s_{A}=ut+\dfrac{1}{2}at^2$

Put the value into the formula

$s_{A}=0+\dfrac{1}{2}\times3\times80$

$s_{A}=120\ m$

Hence, (I). The time elapsed when both particles meet  is 8.9 s.

Nearest option is (2)

(II). The distance covered by A before meeting is 120 m

Nearest option is (3).