Question

A particle is projected vertically upwards with velocity 40 m - Find the ratio of magnitude
of displacement and distance travelled by the particle in 2 s. (Take g = 10 ms?)​

Answer 1

Explanation:

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11th

Physics

Motion in a Straight Line

Distance and Displacement

A particle is projected ver...

PHYSICS

A particle is projected vertically upwards with an initial velocity of 40 m/s. Find the displacement and distance covered by the particle in 6 s. Take g = 10 m/s

2

MEDIUM

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ANSWER

By first equation the maximum height reached

v=u+at

0=40+(−g)(t)

t=

g

40

=4sec

Then, the drop of particle from maximum height

s=ut+

2

1

at

2

[6sec=4sec+2sec]

s=

2

1

(g)(2)

2

=20m

Maximum height reached by particle

s=ut+

2

1

at

2

s=40(4)+

2

1

(10)(4)

2

=40×4+5×16

s=160+80=240m

Thus,

Displacement =240−20=220m

Distance =240+20=260m.

solution

Answer 2

Explanation:

ANSWER

By first equation the maximum height reached

v=u+at

0=40+(−g)(t)

t=g40=4sec

Then, the drop of particle from maximum height

s=ut+21at2              [6sec=4sec+2sec]

s=21(g)(2)2=20m

Maximum height reached by particle

s=ut+21at2

s=40(4)+21(10)(4)2=40×4+5×16

s=160+80=240m

Thus,

Displacement =240−20=220m

Distance =240+20=260m.



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