Question

a body moving with uniform acceleration has velocities 20m/s and 30m/s when passing two points A and B. then the velocity mid way between A and B is approximately
a) 25m/s
b) 25.5. m/s
c) 24 m/s
d) 10m/s

Answer 1

hey friend!!!

In usual notation,

v2 - u2 = 2as = 900 - 400 = 500

Thus, as = 250 --- (1)

At s/2 if speed is w:

w2 - u2 = 2a(s/2)

Which means,

w2 = u2 + as = 400 + 250, using (1).

Or,

w = sqrt(650) = approx 25.5 m/s.

then your answer is option "b"

I hope it will help you..

Answer 2

Answer:

The velocity of body mid way between A and B is 25.5 m/ sec .

Explanation:

• Equation of motion are equations that relates the displacement, velocity, acceleration of a body as a function of time.
• The three equation of motion are:

$v = u + at \\s = ut + \frac{1}{2} a {t}^{2} \\ {v}^{2} - {u}^{2} = 2as$

where, u = initial velocity,

v = final velocity,

s = displacement,

a = acceleration

Given that:

• Initial velocity, u = 20 m/sec
• Final velocity, v = 30 m/ sec
• Let, the displacement between A and B be s meters and the acceleration of the body be a m/sec².
• Using third equation of motion, we get:

${v}^{2} - {u}^{2} = 2as \\ {30}^{2} - {20}^{2} = 2as \\900 - 400 = 2as \\ 500 = 2as \\as = 250 \: \: \: \: \: \: \: \: \: \: \: \: \: - - (1)$

• Let, C be the mid point of A and B. Then, displacement between A and C is s/2 meters.
• Let, the final at point C be v1. Again from third equation of motion, we get:

${v1}^{2} - {u}^{2} = 2a \frac{s}{2} \\ {v1}^{2} - {u}^{2} = as \\ using \: value \: of \: as \: from \: equation \: (1) \\ {v1}^{2} - {20}^{2} = 250 \\ {v1}^{2} - 400 = 250\\ {v1}^{2} = 250 + 400 \\ {v1}^{2} = 650 \\ v1 = \sqrt{650} \\ v1= 25.5$

• Hence, velocity at the mid point between A and B is 25.5 m/sec .