Question

A position–time graph for a particle moving along the $x$ axis is shown in above figure. (a) Find the average velocity in the time interval $t=1.50s$ to $t=4.00s$. (b) Determine the instantaneous velocity at $t=2.00s$ by measuring the slope of the tangent line shown in the graph. (c) At what value of $t$ is the velocity zero.

Answer 1

For average velocity, we find the slope of a secant line running across the graph between the $1.5s$ and $4s$ points. Then for instantaneous velocities we think of slopes of tangent lines, which means the slope of the graph itself at a point. We place two points on the curve: Point $A$, at $t=1.5s$, and Point $B$, at $t=4.0s$, and read the corresponding values of $x$.

(a) At $t_i=1.5s,x_i=8.0m$ (Point A)

At $t_f=4.0s,x_f=2.0m$ (Point B)

$V_{avg}=\frac{x_f-x_i}{t_f-t_i}=\frac{(2.0-8.0)m}{(4.0-1.5)s}$

$-\frac{6.0m}{2.5s}=-2.4m/s$

(b) The slope of the tangent line can be found from points $C$ and $D$.

$(t_C=1.0s,x_C=9.5m)$ and $(t_D=3.5s,x_D=0)$,

$v\approx -3.8m/s$

The negative sign shows that the direction of $v_x$ is along the negative $x$ direction.

(c) The velocity will be zero when the slope of the tangent line is zero. This occurs for the point on the graph where $x$ has its minimum value. This is at $t\approx 4.0s$