displacement, velocity and acceleration questions (1 Viewer)

kissyboy · Jun 19, 2009

the acceleration of a particle moving in a straight line after time t s is given by
a =8-4tms^-2. If the particle started at rest 2m from 0, find:
1) The velocity and displacement as functions of time
2) the time when the particle is stationary and the displacement at this time.
3) the total distance travelled in the first 2 seconds
4) the time interval when the velocity was negative.

Drongoski · Jun 20, 2009

$\80dpi Given:\ \ when\ t=0\ \ \ v=0\ \ \ x=2\ \ \ \ \ a\ =\ \ \ddot{x}\ \ =\ 8-4t\\ \\ \therefore \ v\ =\ \dot{x}\ =\ \int \ddot{x}\ dt\ \ =\ \ \int (8-4t)\ dt\ \ =\ \ 8t-2t^{2}+C\\ \\ When\ t=0,\ v=0\ \Rightarrow \ 8(0)-2(0^{2})+C\ =\ 0\ \Rightarrow \ C=0\\ \\ \therefore \ \ \ v\ =\ \dot{x}\ =\ 8t\ -\ 2t^{2}\\ \\ \therefore \ x\ =\ \int \dot{x}\ dt\ \ =\ \ \int (8t-2t^{2})\ dt\ \ =\ \ 4t^{2}-\frac{2}{3}t^{3}+D\\ \\ When\ t=0,\ x=2\ \Rightarrow \ \ D=2\\ \\ \therefore \ \ x\ =\ 4t^{2}-\frac{2}{3}t^{3}+2\\ \\ (2)\ When\ v=0:\ \ \ \dot{x}\ =\ 8t-2t^{2}\ =\ 2t(4-t)\ =\ 0\ \ \Rightarrow \ \ t= 0,4\ secs\\ \\ \ \ \ x(4)\ =\ 4\times 4^{2}-\frac{2}{3}\times 4^{3}+2\ \ =\ \ 23\frac{1}{3}\ m\\ \\ (3) \ \ x(2)\ =\ 4\times 2^{2}-\frac{2}{3}\times 2^{3}+2\ =\ 7\frac{1}{3}\ m\\ \\ \therefore \ \ total\ distance\ =\ 7\frac{1}{3}\ m\ -\ 2\ m\ \ =\ \ 5\frac{1}{3}\ m\\ \\ (4)\ \ v\ =\ 2t(4-t)\ is\ a\ upside\ down\ parabola\ (quadratic \ in\ t)\\ \\ \Rightarrow\ \ \ v\ negative\ for\ t\ > 4\ secs$

Is this a 2U question ??

Aquawhite · Jun 20, 2009

This appears to be a 2unit question or very easy 3unit question that I've seen before.

Nice answer. Couldn't have worded it any better.

displacement, velocity and acceleration questions (1 Viewer)

kissyboy

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Drongoski

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Aquawhite

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