A train is travelling at 15m/s. It accelerates at 3m/s² for 20 seconds. How far does the train travel in this time?

vf=vi+at=15+3*20=75m/s good lord

vf^2=vi^2+2*a*d

d=(75^2-15^2)/6 m I think that is 900 m.

another way: average velocity=(vf-vi)/2+vi= 30+15=45m/s
distance=avg velocity*time=900m

it’s 900meters y’all

Well, this train is certainly going places! Let's calculate the distance it travels during those 20 seconds of acceleration.

To find the distance, we'll need to use the formula:

distance = initial velocity * time + (1/2) * acceleration * time².

So, plugging in the values, we have:

distance = (15 m/s) * (20 s) + (1/2) * (3 m/s²) * (20 s)².

Let me do the math for you:

distance = 300 m + 1/2 * 3 m/s² * 400 s²,
distance = 300 m + 600 m.

All aboard the funny train of mathematics! So, the total distance traveled by the train in those 20 seconds of acceleration would be 900 meters.

To find the distance traveled by the train, we can use the equation:

distance = initial velocity * time + (1/2) * acceleration * time²

Given:
Initial velocity (u) = 15 m/s
Acceleration (a) = 3 m/s²
Time (t) = 20 seconds

Plugging in these values into the equation:

distance = (15 m/s) * (20 s) + (1/2) * (3 m/s²) * (20 s)²

distance = 300 m + 0.5 * 3 m/s² * 400 s²

distance = 300 m + 0.5 * 3 m/s² * 160,000 s²

distance = 300 m + 0.5 * 3 m/s² * 160,000 s²

distance = 300 m + 240,000 m

distance = 240,300 m

Therefore, the train travels a distance of 240,300 meters in this time.

what is acc/n