Learn interpretation of variable acceleration, practice to calculation of velocity and acceleration, position of particle as a function of time & velocity when acceleration is given in terms of displacement.
A \(a=3\,t^2\)
B \(v=6\,t\)
C \(a=6\,t\)
D \(v=6\,t^2\)
then velocity, \(v=\dfrac{dx}{dt}\)
\(=\dfrac{d}{dt}\,\left[\dfrac{dx}{dt}\right]\)
\(=\dfrac{d^2x}{dt^2}\)
A \(15t^2+3,\;30t\)
B \(16\,t^3+5,\;25t\)
C \(15t^3+3t,\;30t+3\)
D \(5t^2+3,\;15t\)
A \(\dfrac{2}{3}\,sec\,,\;\dfrac{10}{3}\,m\)
B \(\dfrac{2}{5}\,sec\;,\,\dfrac{5}{3}\,m\)
C \(\dfrac{5}{6}\,sec\;,\,\dfrac{12}{5}\,m\)
D \(\dfrac{4}{3}\,sec\,,7\,m\)
A \(-1\,m\,,\,1\,sec\)
B \(-2\,m,\;1\,sec\)
C \(-3\,m,\;1\,sec\)
D \(-4\,m,\;1\,sec\)
Let the position of particle \(=x(t)\)
then velocity, \(v=\dfrac{d[x(t)]}{dt}\) ...(1)
If \(v=\dfrac{d[x(t)]}{dt}\)
\(a=\dfrac{dv}{dt}=\dfrac{d^2}{dt}[x(t)]\) ...(2)
As, \(v=\dfrac{dx}{dt}\)
\(dx=v\,dt\)
Now, integrating both the sides from initial to final conditions, we get
\(\int\limits^x_{x_0}dx=\int\limits ^t_{t_0}v\,dt\)
A \(3t^2-2t+C\)
B \(5t^2-2t+C\)
C \(5t^2+2t+C\)
D \(5t^2+6t+C\)
\(v=\dfrac{dx}{dt}\)
where, \(x\) is the position of the particle.
As, \(v=\dfrac{dx}{dt}\)
Integrating both sides
\(\int\limits ^t_0v\,dt=\int\limits^x_0dx\)
where, \(t\) is the time at which position of the particle is to be determined.
A \(20\,m\)
B \(30\,m\)
C \(40\,m\)
D \(50\,m\)
A \(4t^2+3t+3\)
B \(5t+9\)
C \(4t^2+9t+16\)
D \(8t+3\)
\(\therefore v=f(t)\)
\(a=\dfrac{dv}{dt}\)
where, \(v=\) velocity of the particle
A \(4\,m/sec^2\)
B \(3\,m/sec^2\)
C \(6\,m/sec^2\)
D \(2\,m/sec^2\)
A particle is moving with velocity \(v\), along a straight line given by
\(v=\dfrac{dx}{dt}\)
\(a=\dfrac{dv}{dt}\)
multiplying by \(dx\) at both the sides
\(a.dx=\dfrac{dx}{dt}.dv\)
\(a\,dx=v\,dv\)
\(a=v\,\dfrac{dv}{dx}\) ...(1)
Equation (1) can be described as :
Acceleration as a function of position \(x\).
A \(18x^3+45x^2+25x\)
B \(16x^3+17x^2+20x\)
C \(16x^2+5x+7\)
D \(11x^3+5x\)
\(a=\dfrac{dv}{dt}\)
multiplying both the sides by \(dx\)
\(a\,dx=\dfrac{dx}{dt}.\,dv=v.dv\)
\(a=v\,\dfrac{dv}{dx}\) ...(1)
A \(50\,m/s^2\)
B \(40\,m/s^2\)
C \(88\,m/s^2\)
D \(60\,m/s^2\)
\(a=v\,\dfrac{dv}{dx}\)
\(\int \limits^x_{x_0}a\,dx=\int \limits^v_{v_0}v\,dv\)
A \(3x^2+4x\)
B \(\sqrt{3x^2+4x}\)
C \(x+5\)
D \(2x+5\)
\(a=v\,\dfrac{dv}{dx}\)
\(\int\limits^x_{x_0}a\,dx=\int\limits^v_{v_0}v\,dv\)
A \(54\,m/s\)
B \(\sqrt{27}\,m/s\)
C \(\sqrt{54}\,m/s\)
D \(\sqrt{91}\,m/s\)