Question

1. an olympic - class sprinter starts a race with an acceleration of 4.50 m/s². (a) what is her speed 2.40 s later? (b) sketch a graph of her position vs. time for this period.

Explanation:

Step1: Identify the formula

We use the kinematic - equation $v = v_0+at$. Since the sprinter starts from rest, $v_0 = 0$.

Step2: Substitute the values

Given $a = 4.50\ m/s^{2}$ and $t = 2.40\ s$. Substituting into the formula $v=v_0 + at$, we get $v=0+(4.50\ m/s^{2})\times2.40\ s$.

Step3: Calculate the speed

$v=(4.50\times2.40)\ m/s=10.8\ m/s$.

For part (b), the position - time relationship for an object with constant acceleration starting from rest ($x_0 = 0$ and $v_0 = 0$) is given by the equation $x=\frac{1}{2}at^{2}$. This is a quadratic function of the form $y = Ax^{2}$ where $A=\frac{1}{2}a$. The graph is a parabola opening upwards with the vertex at the origin $(0,0)$. We can calculate some points:
When $t = 0\ s$, $x = 0\ m$; when $t = 1\ s$, $x=\frac{1}{2}\times4.50\ m/s^{2}\times(1\ s)^{2}=2.25\ m$; when $t = 2\ s$, $x=\frac{1}{2}\times4.50\ m/s^{2}\times(2\ s)^{2}=9\ m$; when $t = 2.40\ s$, $x=\frac{1}{2}\times4.50\ m/s^{2}\times(2.40\ s)^{2}=12.96\ m$. Then we can plot these points on a graph with the $x$ - axis representing time $t$ (in seconds) and the $y$ - axis representing position $x$ (in meters) and draw a smooth curve through them.

Answer:

(a) $10.8\ m/s$
(b) The graph of position $x$ vs time $t$ is a parabola opening upwards with the vertex at the origin. Some points on the graph are $(0,0)$, $(1,2.25)$, $(2,9)$, $(2.40,12.96)$ (where the coordinates are $(t,x)$) and a smooth curve is drawn through these points.