a bus is moving on a straight highway at 12m/s. as it approaches the total gate, it decelerates at 2.5 m/s2 in 5 seconds. what is the final velocity of the bus?

Given:

Vi = 12 m/s
a = - 2.5 m/s2
t = 5s
Required: Vf (final velocity)
Solution:
a=Vf^2 - Vi^2 / t
-2.5 m/s = Vf^2 - (12m/s)^2 / 5s
-2.5 m/s = Vf^2 - 144m^2/s^2
-2.5m/s over -144 m^2/s^2 = Vf^2
0.13 m/s = Vf

You must mean toll gate.

After decelerating 5 s at 2.5 m/s^2, it will have reduced its velocity by 12.5 m/s, and will be going backwards, at 0.5 m/s.

A rather strange situation.

To find the final velocity of the bus, we need to use the equation of motion:

final velocity (v) = initial velocity (u) + (acceleration (a) * time (t))

Given:
Initial velocity (u) = 12 m/s
Acceleration (a) = -2.5 m/s^2 (deceleration is considered negative)
Time (t) = 5 seconds

Substituting the values into the equation:

v = u + (a * t)
v = 12 m/s + (-2.5 m/s^2 * 5 s)
v = 12 m/s - 12.5 m/s
v = -0.5 m/s

The final velocity of the bus is -0.5 m/s. However, it's important to note that the negative sign indicates the direction of the velocity change. In this case, the negative sign means the bus is moving in the opposite direction of its initial velocity.