Question

The motion of a particle along a straight line described by the equation x=6+4t2-t4 Where x is in metres(m) and t is the positive time in seconds(s). Find the position, velocity and acceleration at t=2s. During what time interval is the velocity positive During what time interval is x positive What is the maximum possible velocity attained by the particle

Answer 1

Answer:at t=√2 sec velocity is positive and highest

Explanation:for getting velocity differentiate given equation with time and for getting acc. Differentiate velocity with time

Answer 2

Concept:

If the other variables are known, kinematics equations can derive one or more of them. These equations describe motion at a constant velocity or acceleration. We can't apply kinematics equations if one of the two variables is changing.

Find:

The position, velocity and acceleration at the two seconds.

Solution:

$x(t)=6+4t^2-t^4\\x(2)=6+4(2)^2-(2)^4\\x(2)=6+16-16\\x(2)=6m$

The position of the particle is at $6m$ after 2 seconds.

$x(t)=6+4t^2-t^4\\v(t)=8t-4t^3\\v(2)=8(2)-4(2)^3\\v(2)=16-32\\v(2)=-16m/s$

The velocity of the particle is $-16m/s$ after 2 seconds.

$x(t)=6+4t^2-t^4\\v(t)=8t-4t^3\\a(t)=8-12t^2\\a(2)=8-12(2)^2\\a(2)=8-48\\a(2)=-40m/s^2$

The acceleration of the particle is $-40m/s^2$.

$v(t)=8t-4t^3\\\\0=8t-4t^3\\4t^3=8t\\4t^2=8\\t=\sqrt{2}seconds$

The velocity is positive till t reaches $\sqrt{2}$ seconds.

Differentiate the equation of the velocity to find the maximum velocity.

$v(t)=8t-4t^3\\\frac{dv}{dt}=0\\ 8-12t^2=0\\t=\sqrt{\frac{2}{3} } \\v(t)=8t-4t^3\\\\v(\sqrt{\frac{2}{3} } )=8(\sqrt{\frac{2}{3} } )-4(\sqrt{\frac{2}{3} } )^3\\\\v(\sqrt{\frac{2}{3} } )=8(\sqrt{\frac{2}{3} } )-\frac{8}{3} (\sqrt{\frac{2}{3} } )\\v(\sqrt{\frac{2}{3} } )=\frac{16}{3} (\sqrt{\frac{2}{3} } )$

The maximum possible velocity is $\frac{16}{3} (\sqrt{\frac{2}{3} } )$.

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