2. A freight train traveling with a velocity of 18.0 m/s to the south begins braking as it approaches the train yard. The train’s acceleration while braking is -0.33m/s2. What is the train’s speed after 23 seconds?
I have the same problem!! I'm not sure but I think you would multiply 23 times -.33 to find how much the speed is changing in 23 seconds. That is -7.59 so therefore the speed, over any 23 seconds is decreasing 7.59 meters per second. Then I subtracted 7.59 m/s from 18 m/s to find the trains speed after 23 seconds. That is 10.41 m/s, and that was my answer.
literally 2022 for me
dude. WHAT ARE THE ODDS OF ME SEEING THIS EXACT PROBLEM IN 2020!? I REALLY hope your right. GOD! WORD PROBLEMS!
WHY DO THEY KEEP CHANGING MATH AND ALGERBRA?!?!?!?!?!?
I hope my mom doesn't see this...OH CRAP! GTG!
To find the train's speed after 23 seconds, we need to use the equation of motion that relates velocity (V), initial velocity (U), acceleration (a), and time (t):
V = U + at
Given:
Initial velocity (U) = 18.0 m/s (south)
Acceleration (a) = -0.33 m/s^2 (negative sign indicates deceleration)
Time (t) = 23 seconds
We will assume that the positive direction is towards the south.
Substituting the given values into the equation, we have:
V = 18.0 m/s + (-0.33 m/s^2) * 23 s
To calculate the final speed, we multiply the acceleration by time and subtract it from the initial velocity.
Let's calculate the value of V.