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2 years ago

6

17. How long will it take for an object accelerating at a constant rate of 5 m/s to change its velocity from 0 to 6 m/s? (A) 0.6

s (B) 1.2 s (C) 2.4 s (D) 3.6 s (E) 4.8 s

1 answer:

VikaD [51]2 years ago

8 0

Answer:

(B)

Explanation:

Time = change of velocity ÷ acceleration

= (6-0) ÷ 5

= 1.2

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A rocket is fired vertically upwards with initial velocity 92 m/s at the ground level. Its engines then fire and it is accelerat

Readme [11.4K]

Answer:

The rocket above the ground is in 44 sec.

Explanation:

Given that,

Initial velocity = 92 m/s

Acceleration = 4 m/s²

Altitude = 1200 m

Suppose, How long was the rocket above the ground?

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Using equation of motion

$s=ut-\dfrac{1}{2}at^2$

Put the value into the formula

$1200=92t+\dfrac{1}{2}\times4t^2$

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$v=u+at$

Put the value into the formula

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When the rocket hits the ground,

Then, h'=0

We need to calculate the time

Using equation of motion

$h'=h+ut-\dfrac{1}{2}at^2$

Put the value into the formula

$0=1200+132t-\dfrac{1}{2}\times9.8t^2$

$4.9t^2-132t-1200=0$

$t=34\ sec$

When the rocket is in the air it is the sum of the time when it reaches 1000 m and the time when it hits the ground

So, the total time will be

$t'=34+10$

$t'=44\ sec$

Hence, The rocket above the ground is in 44 sec.

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