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Question

Physics Question on Acceleration

A particle moves along Y-axis in such a way that its yy- coordinate varies with time tt according to the relation y=3+5t+7t2.y=3+5t+7{{t}^{2}}. The initial velocity and acceleration of the particle are respectively:

A

14ms1,5ms214m{{s}^{-1}},-5\,m{{s}^{-2}}

B

19ms1,9ms219\,m{{s}^{-1}},-9\,m{{s}^{-2}}

C

14ms1,5ms2-14\,m{{s}^{-1}},-5\,m{{s}^{-2}}

D

5ms1,14ms25m{{s}^{-1}},14m{{s}^{-2}}

Answer

5ms1,14ms25m{{s}^{-1}},14m{{s}^{-2}}

Explanation

Solution

Rate of change of position is known as velocity and rate of change of velocity is known as acceleration.
Given y=3+5t+7t2y=3+5t+7{{t}^{2}}
Velocity (v)(v) is defined as rate of change of displacement, also using
ddxxn=nxn1,\frac{d}{dx}{{x}^{n}}=n{{x}^{n-1}},
we have \therefore v=dydt=5+14tv=\frac{dy}{dt}=5+14\,t
Initial velocity at t=0,t=0, is v=5m/s.v=5\,m/s.
Also acceleration a=d2ydt2=14m/s2a=\frac{{{d}^{2}}y}{d{{t}^{2}}}=14\,m/{{s}^{2}}