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Question

Physics Question on Motion in a straight line

A particle moves along Y-axis in such a way that its y-coordinate varies with time tt according to the relation y=3+5t+7t2y = 3 +5t + 7t ^2. The initial velocity and acceleration of the particle are respectively

A

14m14 m s1,5ms2s^{-1}, -5m\,s^{-2}

B

19m19 m s1s^{-1},9m-9\,m s2s^{-2}

C

14m-14\, m s1s^{-1},5m-5\,m s2s^{-2}

D

5m5 \,m s1s^{-1},14m14\,m s2s^{-2}

Answer

5m5 \,m s1s^{-1},14m14\,m s2s^{-2}

Explanation

Solution

Given y=3+5t+7t2y = 3 + 5t+7t^2
Velocity (v) is defined as rate of change of displacement, also
using ddxxn=nxn1,\frac{d}{dx}x^n=nx^{n-1}, we have
hence v=dydt=5+15tv=\frac{dy}{dt}=5+15t
Initial velocity at t = 0, is
v=5ms1v=5ms^{-1}
Also acceleration a =d2ydt2\frac{d^2 y}{dt^2}
=14ms2=14 ms^{-2}