Question

A particle has an initial velocity of 31 +49 and an acceleration of 0.41 +0.3). The magnitude of its velocity
after 10 s is
SOLVE IT BY STEP BY STEP. .........​

Answer 1

Answer:

87.1units

Explanation:

INITIAL VELOCITY OF PARTICLE= 31+49= u=80UNITS

ACCELERATION=0.41+0.3= 0.71 UNITS

SINCE PARTICLE IS MOVING WITH INITIAL VELOCITY NOT EQUALS TO 0.

THUS AFTER 10SECS IT'S FINAL VELOCITY WILL BE = v

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FROM UPPER THINGS KEYWORDS WE HAVE:

1. INITIAL VELOCITY = u = 80 units
2. ACCELERATION= 0.71 units
3. TIME= 10 secs
4. FINAL VELOCITY= v = ?

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FIRST EQUATION OF MOTION SAYS THAT -

FINAL VELOCITY OF A BODY IS EQUAL TO SUM OF ITS INITIAL VELOCITY AND PRODUCT OF ITS ACCELERATION WITH THE TIME IT'S BEING ACCELERATED.

THUS WE CAN SAY:

v = u + at -----------------(FIRST EQUATION OF MOTION)

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AFTER Inputting THE VALUES IN THIS EQUATION WE GET:

v = 80 + ( 0.71 ×10 )

$v = 80 + (7.1) \\ v = 80 + 7.1 \\ v = 87.1 \: units$

THEREFORE THE FINAL VELOCITY OF THE PARTICLE WILL BE 87.1 UNITS.

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BY SCIVIBHANSHU

THANK YOU

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