Question

a uniformly accelerated body passes two pt P and Q with speed of 10 m/s & 20 m/s .if o is the mid pt of P and
Q

Answer 1

Topic :- Motion in staright line

$\maltese\:\underline{\sf Complete \: Question :}\:\maltese$

A uniformly accelerated body passes two pt P and Q with speed of 10 m/s & 20 m/s. If O is the midpoint of P and Q then speed of O will be ?

$\maltese\:\underline{\sf AnsWer :}\:\maltese$

If any body is moving with an uniform Acceleration from point P to Q along a staright line having velocities $\sf v_1$ and $\sf v_2$ at point P and Q respectively. If O is the midpoint between P and Q then Velocity of the body at point will be calculated by given below formula :

$:\implies \sf v = \sqrt{\dfrac{v_1^{2} + v_2^{2} }{2}}$

Here,

$\bullet\:\underline{\sf v_1 = 10 \: m/s} \\ \\ \bullet\:\underline{\sf v_2 = 20 \: m/s}$

Now, simply substitue this given values in above formula :

$:\implies \sf v = \sqrt{\dfrac{ {(10)}^{2} + {(20)}^{2} }{2}}$

$:\implies \sf v = \sqrt{\dfrac{ 100 + 400 }{2}}$

$:\implies \sf v = \sqrt{\dfrac{500 }{2}}$

$:\implies \sf v = \sqrt{250}$

$:\implies \sf v = \sqrt{5 \times 5 \times 10}$

$:\implies \sf v =5 \sqrt{10}$

$:\implies \sf v =5 \times 3.16$

$:\implies \underline{ \boxed{\sf v =15.8 \: m {s}^{ - 1}}}$

Hence,the speed of the object at point O will be 15.8 m/s.

Answer 2

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♧Correct Question♧

• A uniformly accelerated body passes two pt P and Q with speed of 10 m/s & 20 m/s .if o is the mid pt of P and Q.then speed of O will be____.

♧Answer♧

Given data:-

• velocity = V1 = 10m/s
• velocity = V2 = 20m/s

To find:-

• speed = v = ?

Solution:-

☆P and Q have velocities V1 and V2 at point P and Q respectively at mid point O.

By using this formula,

v = √((V1² + V2²)/2)

• Put values in the formula,
• ➠v = √((10² + 20²)/2)
• ➠v = √((100 + 400)/2)
• ➠v = √(500/2)
• ➠v = √(250)
• ➠v = 15.8m/s. Ans

☆So the speed will be 15.8m/s at O.

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