Question

A particle has displacement equation
(i) XA= 2t+7
(ii) Xb=3t2+2t+6
(iii)xc= 5ť +4t, which of them has uniform acceleration?​

Answer 1

Answer:

In this type of question the displacement function w.r.t. time is given. You have to differentiate it twice to get acceleration function . If the acceleration is numerically an integer, then it is uniform.

1st case.

XA = 2t + 7

=> d (XA)/dt = VA =2

=> d(VA)/dt = 0, therefore acceleration in this case is zero (0).

2nd case.

XB = 3t^2 +2t + 6

=> d(XB)/dt = VB = 6t +2

=> d(VB)/dt = 6

therefore the acceleration in this case is 6 m/s^2 : uniform acceleration

3rd case.

XC = 5t^3 + 4t

=> d(XC)/dt = VC = 15t^2 + 4

=> d(VC)/dt = 30 t , therefore acceleration in this case is time dependent and hence not uniform.

Answer 2

Answer:

$\bold\red{(ii) X_{b}=3{t}^{2}+2t+6}$

Explanation:

Given,

A particle having displacement terms of t.

Now,

we know that,

we have to differentiate twice the displacement function wrt time to get the acceleration.

(i) X = 2t + 7

Differentiating both sides wrt t,

we get,

$= > \frac{dx}{dt} = 2 \\ \\ = > \frac{ {d}^{2}x }{d {t}^{2} } = 0$

Here we get,

acceleration = 0

$(ii) \: x = 3 {t}^{2} + 2t + 6$

Differentiating both sides wrt t,

we get,

$= > \frac{dx}{dt} = 6t + 2 \\ \\ = > \frac{ {d}^{2} x}{d {t}^{2} } = 6$

Here,

acceleration = 6

Clearly,

It is constant

$(iii)X = 5{t}^{3}+4t$

Differentiating both sides wrt t,

we get,

$= > \frac{dx}{dt} = 15{t}^{2}+4\\ \\ = > \frac{ {d}^{2} x}{d {t}^{2} } = 30t$

Here,

acceleration = 30t

It's dependent on time

Hence,

Correct Option is

$\bold{X_{b}=3{t}^{2}+2t+6}$