Question

If displacement of a particle is directly proportional to the square of time. Then particle is moving with
(a) Uniform acceleration
(b)
Variable acceleration
(c) Uniform velocity
(d) Variable acceleration but uniform velocity​

Answer 1

Answer:

a) uniform acceleration

Explanation:

$s \: \alpha \: {t}^{2} \\ \\ \frac{ds}{dt} \: \alpha \: \: 2{t}^{} \\ \\ v \: \alpha \: t \\ \\ \frac{dv}{dt} \: \alpha 1 \\ \\ a = constant$

Note:-

• a = dv/dt
• v = ds/dt
• d(x)^n/dx = nx^n-1
• d(x)/dx = 1

Also note:-

• a = acceleration
• v = velocity
• s = displacement

Answer 2

Answer:

• Uniform acceleration :- If the rate if change of velocity with respect to time is constant then it is called uniform acceleration.

Explanation:

Let's consider the given statement that

Displacement is proportional to square of time

Here , I consider the following variables as following symbols

• a = acceleration
• v = velocity
• s = displacement

$s \: \alpha \: {t}^{2} \\ \\ \dfrac{ds}{dt} \: \alpha \: \: 2{t}^{} \\ \\ v \: \alpha \: t \\ \\ \dfrac{dv}{dt} \: \alpha 1 \\ \\ a = constant$

Note:-

• a = $\dfrac{dv}{dt}$
• v = $\dfrac{ds}{dt}$
• $\dfrac{d{x}^{n}}{dt}=n{x}^{n-1}$= nx^n-1
• $\dfrac{dx}{dx}=1$

So your required answer is Uniform acceleration