Abird, accelerating from rest at a constant rate, experiences a displacement of 28 m in 11 s. what is the final velocity after 11 s?

Let's see what variables we've got first. Hmmm. We have:

Displacement, d = 28 m
Time taken, t = 11 s
Initial velocity, u = 0 m/s (at rest)

And now we need to find the final velocity, v. Among the 4 (or 5) equations of motions, there's no equation that will let us simply plug in the values and give an answer sigh. But fear not! We'll do it in steps.

I'm going to pick one of the motion equation to find more information:



I know everything except for a in this one, so I I'll use this! After plugging in values, I get a = 0.4628 m/s^2.

Now I'm going to use another motion equation that has v in it because that needs to be solved!



Now I know everything except dial velocity v. Nice!

v = 0 + (0.4628)(11)

Since you have not included the experiment situation and the data, of course, i can not give you specific details about it. but i can you with this question.1) i will assume that the motion is projectile (parabolic) motion.2) yes, the horizontal distance depends on the horizontal velocity as, naming vx the horizontal velocity and t the time, it is: x = vx × t3) now, observe that the flight time depends on the vertical velocity as, naming voy the initial vertical velocity, t the time and y the vertical distance before reaching the ground: y = yo + voy×t - gt² / 2.the higher the initial vertical velocity (voy) the longer the object will be flying and so t will be higher in the equation for the horizontal distance.that explains the relationship between the horizontal distance and the horizontal velocity.