Question

the relation between position(x) and time(t) are given below for a particle moving along a straight line. which of the following equation represents uniformly accelerated motion? [ where a and b are positive contants].

1) bx = at + ab

2) ax = b + t

3) xt = ab

4) at = (b+x)1/2

Answer 1

When position is given as a function of time, the velocity and acceleration are given as follows:

$v = \frac{dx}{dt} \\ \\ a = \frac{dv}{dt}$


Now, we have to find which of the following options have uniformly accelerated motion.


That is, we have to find in which option the following condition is satisfied:

$a = constant \\ \\ OR \quad \frac{dv}{dt} = constant \\ \\ OR \quad \frac{d^2x}{dt^2} = constant$

[Note that the constant must not be zero]

____________________________

We can now check each option individually.

$a$ is already used as a constant. So we will denote acceleration by $f$.




$1) \, \, bx = at + ab \\ \\ \\ \text{Differentiating w.r.t time} \\ \\ \\ \implies b \frac{dx}{dt} = a + 0 \\ \\ \\ \implies bv = a \\ \\ \\ \text{Differentiating again w.r.t time} \\ \\ \\ \implies b\frac{dv}{dt} = 0 \\ \\ \\ \implies bf = 0 \\ \\ \\ \implies \boxed{\bold{f=0}}$

The acceleration is zero. So, Option (1) does not represent a Uniformly Accelerated Motion.




$2) \, \, ax = b+t \\ \\ \\ \text{Differentiating w.r.t time} \\ \\ \\ \implies a \frac{dx}{dt} = 0+1 \\ \\ \\ \implies av = 1 \\ \\ \\ \text{Differentiating again w.r.t time} \\ \\ \\ \implies a\frac{dv}{dt} = 0 \\ \\ \\ \implies af = 0 \\ \\ \\ \implies \boxed{\bold{f=0}}$

Again, the acceleration is zero. So, Option (2) also does not represent a Uniformly Accelerated Motion.




$3) \, \, xt = ab \\ \\ \\ \implies x = \frac{ab}{t} \\ \\ \\ \text{Differentiating w.r.t time} \\ \\ \\ \implies \frac{dx}{dt} = -\frac{ab}{t^2} \\ \\ \\ \implies v = -\frac{ab}{t^2}\\ \\ \\ \text{Differentiating again w.r.t time} \\ \\ \\ \frac{dv}{dt} = \frac{2ab}{t^3} \\ \\ \\ \implies \boxed{\bold{f = \frac{2ab}{t^3}}}$

Here we see that Acceleration is Inversely Proportional to the cube of time. Clearly, it is not constant. It is dependent on time. So, Option (3) also does not represent a Uniformly Accelerated Motion



$4) \, \, at = \left( b+x \right)^\frac{1}{2} \\ \\ \\ \text{Squaring both sides} \\ \\ \\ \implies a^2t^2 = b+x \\ \\ \\ \text{Differentiating w.r.t time} \\ \\ \\ \implies a^2(2t) = 0 + \frac{dx}{dt} \\ \\ \\ \implies 2a^2t = v \\ \\ \\ \text{Differentiating again w.r.t time} \\ \\ \\ \implies 2a^2 (1) = \frac{dv}{dt} \\ \\ \\ \implies \boxed{\bold{f=2a^2=constant}}$


Finally, we see that in this option, acceleration does come out to be constant. Thus, Option (4) represents a Uniformly Accelerated Motion.




Thus, The Final Answer is Option (4) $\bold{at = \left(b+x\right)^\frac{1}{2}}$

Answer 2

Answer:

option d

Explanation: