An MRT Train accelerates at a rate of 2.75 m/s^2 from rest for 3.00 seconds.How fast will the train be moving after that time?

v = V0 + a * t

V0 = 0
a = 2.75 m/s^2
t = 3.00 s
v = 0 + 2.75 * 3.00 m/s
v = 8.25 m/s

Well, it sounds like the train is in a hurry to get to its destination! Let's do some calculations to figure out its speed.

Since the train starts from rest, its initial velocity is 0 m/s.

Using the formula for acceleration, which is a = (vf - vi) / t, we can rearrange it to solve for the final velocity, vf.

a = (vf - vi) / t

Rearranging gives us:

vf = (a * t) + vi

Plugging in the given values:

vf = (2.75 m/s^2 * 3.00 s) + 0 m/s

Calculating:

vf = 8.25 m/s

So, after 3.00 seconds, the MRT train will be moving at a speedy 8.25 m/s! Watch out for speedy trains, they might just beat you to the station!

To determine how fast the train will be moving after 3.00 seconds of acceleration, we can use the kinematic equation:

v = u + at

Where:
v = final velocity (unknown)
u = initial velocity (0 m/s, because the train starts from rest)
a = acceleration (2.75 m/s^2)
t = time (3.00 seconds)

Plugging the given values into the equation:

v = 0 + (2.75 m/s^2)(3.00 s)

Simplifying:

v = 8.25 m/s

Therefore, the train will be moving at a speed of 8.25 m/s after 3.00 seconds of acceleration.

To find the final velocity of the train after accelerating, we can use the equation:

v = u + at

where:
v = final velocity
u = initial velocity (which is 0 m/s as the train starts from rest)
a = acceleration
t = time

Plugging in the given values:

u = 0 m/s (initial velocity)
a = 2.75 m/s^2 (acceleration)
t = 3.00 seconds (time)

v = 0 + (2.75 × 3.00)
v = 0 + 8.25
v = 8.25 m/s

Therefore, the train will be moving at a speed of 8.25 m/s after 3.00 seconds of acceleration.