A particle has displacement equation
(i) XA= 2t+7
(ii) Xb=3t2+2t+6
(iii)xc= 5ť +4t, which of them has uniform acceleration?
Answers
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:
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,
Here we get,
acceleration = 0
Differentiating both sides wrt t,
we get,
Here,
acceleration = 6
Clearly,
It is constant
Differentiating both sides wrt t,
we get,
Here,
acceleration = 30t
It's dependent on time
Hence,
Correct Option is