A train is traveling down a straight track at 22 m/s when the engineer applies the brakes, resulting in an acceleration of -1.0 m/s2 as long as the train is in motion. How far does the train move during a 40 s time interval starting at the instant the brakes are applied?
Distance = velocity(initial) * time + 1/2 * acceleration * time^2
D = 22 * 40 + 1/2 * (-1) * 1600
D = 880 + (-800)
D = 80
What's wrong with my approach?
Thanks.
You used the equation for displacement, not distance.
Here is a clue: what is final velocity at t=0?
vf=22 -1*40=22-40= -18/s, so it is moving back toward the origin.
What was average velocity (22+(-18))/2=
= 2m/s
What displacement does it travel in 40 sec?
d= avg velocity*time=80m
Now, what distance did it travel?
distance= 22+18/2=20m/s x 40 sec=800m
I see my mistakes but the homework site doesn't accept 800m.
Ana: I reread the dang problem. Key phrase: " as long a the train in motion" For some of the time of 40 seconds, the train is stopped.
Vf^2=Vi^2+2ad
0=22^2-2d
d= 22^2/2 m
Your initial approach is correct in using the formula for distance, but there seems to be an error in the calculation for the acceleration term.
In the given problem, the acceleration is stated as -1.0 m/s^2 when the train is in motion. However, this acceleration is not constant and only applies as long as the train is in motion. Once the train comes to a stop, the acceleration would be zero.
To calculate the distance traveled, you need to split the time interval into two parts: the first part when the train is decelerating and the second part when the train has already stopped.
First, let's calculate the distance covered during the deceleration phase:
Distance1 = initial velocity * time1 + 1/2 * acceleration * time1^2
In this case, the initial velocity is 22 m/s, the time is the duration of the deceleration phase (let's call it t1), and the acceleration is -1.0 m/s^2.
Distance1 = 22 * t1 + 1/2 * (-1) * t1^2
Now, let's calculate the time t1 when the train comes to a stop:
0 = 22 + (-1) * t1
t1 = 22
Now that we know the deceleration time t1, we can calculate the distance covered during this phase:
Distance1 = 22 * 22 + 1/2 * (-1) * 22^2
Distance1 = 484 - 242
Distance1 = 242 m
Next, let's calculate the distance covered in the second phase where the train has already stopped. The time interval for this phase is given as 40 seconds, and there is no acceleration since the train has stopped.
Distance2 = initial velocity * time2
In this case, the initial velocity is 0 m/s, and the time is the remaining time after the deceleration phase.
time2 = 40 - t1
time2 = 40 - 22
time2 = 18
Distance2 = 0 * 18
Distance2 = 0
Finally, we can calculate the total distance traveled by the train during the 40-second time interval:
Total Distance = Distance1 + Distance2
Total Distance = 242 + 0
Total Distance = 242 m
So the correct answer is 242 meters.