Question

body moving with a uniform acceleratin had.
velocities of 20 m/s and 30m/s when passing
the points P and Q of its path. Find the ve-
locity midway between P and Q in m/s)
400+900
1''
Y
2
= 1650​

Answer 1

You might've guessed the answer as 25 m/s, but that's incorrect.

By uniform acceleration, we mean that the acceleration is constant with time. So, the answer would've been 25 m/s if we were asked the velocity mid-time in crossing P and Q.

Anyways, to solve:

Let acceleration be a, and the distance between P and Q be d.

So, $30^2 = 20^2 + 2ad$ using equation of motion

Solving, we get: ad = 250.

Now, when calculating the velocity at the point midway, we'll take the distance to be d/2, with the same acceleration, and initial velocity 20 m/s (or final 30 m/s, but that one was simpler).

Applying the equation of motion:

$v^2 = 20^2 + 2a\frac{d}{2} = 400 + ad$

Using ad = 250:

$v^2 = 400 + 250 = 650$

And, v = $\sqrt{650}$ m/s, which is a scary number.

Properly, v = 25.495 m/s.

You'd probably say that HEY THAT'S PRETTY MUCH 25, but nah, wrong concept :3

Answer 2

Explanation:

Let acceleration be a, and the distance between P and Q be d.

So,  using equation of motion

Solving, we get: ad = 250.

Now, when calculating the velocity at the point midway, we'll take the distance to be d/2, with the same acceleration, and initial velocity 20 m/s (or final 30 m/s, but that one was simpler).

Applying the equation of motion:

Using ad = 250:

And, v = $\sqrt{650}$ m/s, which is a scary number.

Properly, v = 25.495 m/s.