Question

1. (35 points) a train accelerates uniformly from rest to a velocity of 32.5 m/s. during that time the train travels 12500 meters. how long did this motion take?

Explanation:

Step1: Identify the kinematic - equation

We use the equation $x = v_0t+\frac{1}{2}at^{2}$ and $v = v_0 + at$. Since $v_0 = 0$ (starts from rest), $x=\frac{1}{2}at^{2}$ and $v = at$. First, from $v = at$, we can express $a=\frac{v}{t}$. Substitute $a$ into $x=\frac{1}{2}at^{2}$, we get $x=\frac{1}{2}\times\frac{v}{t}\times t^{2}=\frac{1}{2}vt$.

Step2: Rearrange the formula for time

We know that $x = 12500$ m and $v = 32.5$ m/s. Rearranging the formula $x=\frac{1}{2}vt$ for $t$, we have $t=\frac{2x}{v}$.

Step3: Calculate the time

Substitute $x = 12500$ m and $v = 32.5$ m/s into the formula $t=\frac{2x}{v}$. So $t=\frac{2\times12500}{32.5}=\frac{25000}{32.5}\approx80$ s.

Answer:

80 s