Question

question 15
5 points save answer
a train with a constant speed of 16 m/s passes through a town. after leaving the town, the train accelerates at 0.33 m/s² until it reaches a speed of 35 m/s. how far did the train travel while it was accelerating?
3.0 km
1.5 km
2.3 km
0.53 km
0.029 km

Explanation:

Step1: Identify the kinematic - equation

We use the equation $v^{2}=v_{0}^{2}+2ax$, where $v$ is the final velocity, $v_{0}$ is the initial velocity, $a$ is the acceleration, and $x$ is the displacement.

Step2: Rearrange the equation for $x$

$x=\frac{v^{2}-v_{0}^{2}}{2a}$

Step3: Substitute the given values

Given $v_{0} = 16\ m/s$, $v = 35\ m/s$, and $a=0.33\ m/s^{2}$.
$x=\frac{35^{2}-16^{2}}{2\times0.33}=\frac{(35 + 16)(35 - 16)}{0.66}=\frac{(51)\times(19)}{0.66}=\frac{969}{0.66}\approx1468.18\ m$

Step4: Convert to kilometers

$x = 1468.18\ m=1.46818\ km\approx1.5\ km$

Answer:

1.5 km