The displacement of a particle after time $t$ is given by $x = \left( {k/{b^2}} \right)\left( {1 – {e^{ – bt}}} \right)$ where $b$ is a constant. What is the acceleration of the particle?
The acceleration of a moving body can be found from
Figure gives the $x -t$ plot of a particle executing one-dimensional simple harmontc motion. Give the signs of position, velocity and acceleration variables of the particle at $t=0.3 \;s , 1.2\; s ,-1.2\; s$
Draw $x \to t$ graph for zero acceleration.
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$x=\frac{1}{2} a t^2 \quad$ (if coefficients of $t^2$ is positive $(a > 0)$ upward opening parabola) 