A train is speeding along a straight, level track at +20.0 m/s when the engineer applies the brakes, which results in an acceleration of -0.894 m/s2 as long as the train is in motion. How far does the train move during the 70 sec after the brakes are first applied?
Key phrase is “as long as the train is in motion”: it moves during the time interval t<70 s.
The train stops after t seconds of braking
t=(v-vₒ)/a=( 0-20)/(-0.894)=22.4 s.
s= vₒt-at²/2 =20•22.4 – 0.894•22.4²/2=223.7 m
To find the distance the train moves during the 70 seconds after the brakes are first applied, we can use the kinematic equation:
distance = initial velocity * time + (1/2) * acceleration * time^2
In this case, the initial velocity is +20.0 m/s (positive because it is in the direction of motion), the acceleration is -0.894 m/s² (negative because it opposes the motion), and the time is 70 seconds.
Plugging in the values, we have:
distance = 20.0 m/s * 70 s + (1/2) * (-0.894 m/s²) * (70 s)^2
Now we can calculate the distance.