As a train accelerates uniformly, it passes successive 5-kilometer markers while traveling at velocities 10 m/s and 25 m/s. what is the velocity in km/h when it passes the next marker? round to nearest tenth of a km/h.

122.0 km/hr. First let’s make sure all of our units are in the base meter form: i. e. convert 5km to 5000m. (We will convert back to km later). The first thing to do is look at the equation relating velocity, acceleration, and distance: Vf^2 = Vi^2 + 2*a*d, where Vf is final velocity, Vi is initial velocity, a is acceleration, and d is distance. 25^2 = 10^2 + 2*a*5000 = ?> 625 = 100 + 10000a = > a = 0.0525m/s^2. Now that we have acceleration, we can use the same equation again with different numbers.: Vf^2 = Vi^2 + 2*a*d = 25^2 + 2*0. 0525m*5000 = 625 + 525 = 1150 = > Vf^2 = 1150 = > 33.9m/s. Convert to km/hour: 33.9m/s * 1km/1000m * 60s/1min * 60min / 1 hr = 122.0 km/hr.