Question

an antelope moving with constant acceleration covers the distance 75.0 m between two points in time 6.50 s. its speed as it passes the second point is 14.0 m/s. part a what is its speed at the first point? express your answer with the appropriate units. part b what is the acceleration? express your answer with the appropriate units.

Explanation:

Step1: Use average - speed formula

The average speed formula is $v_{avg}=\frac{d}{t}$, where $d = 75.0$ m and $t=6.50$ s. So, $v_{avg}=\frac{75.0}{6.50}\text{ m/s}\approx11.54$ m/s. Also, the formula for average speed in terms of initial speed $v_0$ and final speed $v$ is $v_{avg}=\frac{v_0 + v}{2}$, with $v = 14.0$ m/s.

Step2: Solve for initial speed $v_0$

From $v_{avg}=\frac{v_0 + v}{2}$, we can re - arrange to get $v_0=2v_{avg}-v$. Substituting $v_{avg}\approx11.54$ m/s and $v = 14.0$ m/s, we have $v_0=2\times11.54 - 14.0=23.08 - 14.0 = 9.08$ m/s.

Step3: Use kinematic equation to find acceleration

We use the kinematic equation $v=v_0+at$. We know $v = 14.0$ m/s, $v_0 = 9.08$ m/s and $t = 6.50$ s. Rearranging for $a$, we get $a=\frac{v - v_0}{t}$.

Step4: Calculate acceleration

Substitute the values into the acceleration formula: $a=\frac{14.0 - 9.08}{6.50}=\frac{4.92}{6.50}\text{ m/s}^2\approx0.757\text{ m/s}^2$.

Answer:

Part A:
$v = 9.08$ m/s
Part B:
$a=0.757$ m/s$^2$