Name: Motion With Variable Acceleration
Uploaded: 2019-07-19
Duration: 10 min 54 s
Description: You may be aware of the SUVAT equation, but do you think this can be applicable for all types of motion? No, what are those motion? Watch this to find out.

Question

In the figure shown, $A$ and $B$ are two particles which start from rest. $A$ has constant acceleration $a$ in the direction shown. $B$ also increases its speed at a constant rate $b$ but the direction of velocity is always towards $A$ . Find the time after which $B$ meets $A$ . Also, find the total distance travelled by $B$ $(b > a)$ .

Accepted Answer

Given,
initial velocities of $B$ $= 0$ ,
acceleration of $A$ along line $A A^{'} = a$ ,
magnitude of acceleration of $B = b$ ,
distance between points $A and B = l$ .

Initially, the particle $B$ is at point $B$ and it moves in the direction of $B B^{'}$ . The particle $A$ is at point $A$ initially. After some time, particle $B$ is at position $B$ and particle $A$ is at position $A$ . It is given that the particle $B$ is always moving towards particle $A$ . Suppose at this time, the angle made by velocity of particle $B$ with the direction of $B B^{'}$ is $θ$ and the total time after $B$ will catch particle $A$ is $T$ .
Assume, $A A^{'}$ is along $x$ -direction.
The distance travelled by $B$ in $x$ -direction in $T$ time will be equal to distance travelled by particle $A$ in $T$ time.
$∴ \int_{0}^{T} v_{b} \cos (θ) d t = \frac{1}{2} a T^{2} \dots (i)$
$B$ are velocities of $A$ and $B$ at any time $t$ and their value will be, $v_{a} = a t$ and $v_{b} = b t$ .
Put this value in equation $(i)$ ,
$\Rightarrow \int_{0}^{T} b t \cos (θ) d t = \frac{1}{2} a T^{2}$
$\Rightarrow \int_{0}^{T} t \cos (θ) d t = \frac{a}{2 b} T^{2} \dots (ii)$

The relative velocity of approach of $B$ to $A$ at instant $t$ will be,
$\Rightarrow V = v_{b} - v_{a} \cos (θ)$
$\Rightarrow V = b t - a t \cos (θ)$ .

Since the velocity is given by the rate of change of displacement, hence,
$- V = \frac{d r_{AB}}{d t}$ (negative sign because the distance between them is decreasing)

$\Rightarrow d r_{AB} = - V d t$ .

Integrating both sides,

$\Rightarrow \int_{l}^{0} d r_{AB} = - \int_{0}^{T} (b t - a t \cos (θ)) d t$
$\Rightarrow - l = - \int_{0}^{T} b t d t - \int_{0}^{T} a t \cos (θ) d t$
$\Rightarrow l = \frac{b T^{2}}{2} - a \int_{0}^{T} t \cos (θ) d t$ .

Putting value of $\int_{0}^{T} t \cos (θ) d t$ from equation $(ii)$ ,
$\Rightarrow l = \frac{b T^{2}}{2} - a (\frac{a}{2 b} T^{2})$
$\Rightarrow T = \sqrt{\frac{2 l b}{b^{2} - a^{2}}}$ .

Therefore, the total time required for particle $B$ to meet with particle $A$ will be $\sqrt{\frac{2 l b}{b^{2} - a^{2}}}$ .
Since the magnitude of velocity of particle $B$ is increasing at rate $b$ , hence, the distance $(s)$ travelled by it in time $T$ will be,
$\Rightarrow s = \frac{1}{2} b T^{2}$ .

Putting the value of $T$ in the above equation, $\Rightarrow s = \frac{1}{2} b (\frac{2 l b}{b^{2} - a^{2}}) = \frac{b^{2} l}{b^{2} - a^{2}}$ .

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