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Tresset [83]
4 years ago
15

A particle moves in the xy plane with constant acceleration. At time zero, the particle is at x = 6 m, y = 8.5 m, and has veloci

ty ~vo = (9 m/s) ˆı + (−2.5 m/s) ˆ . The acceleration is given by ~a = (4.5 m/s 2 ) ˆı + (3 m/s 2 ) ˆ . What is the x component of velocity after 3.5 s? Answer in units of m/s.
Physics
1 answer:
andreev551 [17]4 years ago
8 0

Answer:

24.75 m/s

Explanation:

Since it's only asking for the component we can use v_final=v_initial+at which then gives v_final=9+3.5*4.5=24.75 m/s

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

We need to calculate the time

Using equation of motion

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

Put the value into the formula

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

2t^2+92-1200=0

t=10\ sec

We need to calculate the velocity

Using equation of motion

v=u+at

Put the value into the formula

v=92+4\times10

v=132\ m/s

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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Starting with a constant velocity of 45 km/h, a car accelerates for 35 seconds at an acceleration of 0.45 m/s2 . What is the vel
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Explanation:

Vi = 45 Km/h = 12.5 m/s

Vf - Vi = at

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Vf= 28.3 m/s

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