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3 years ago

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45. A particle moves in a straight line with an initial velocity of 30 m/s and constant acceleration 30 m/s2 . (a) What is its d

isplacement at t = 5 s? (b) What is its velocity at this same time?

2 answers:

omeli [17]3 years ago

7 0

Answer:

Displacement after 5 seconds is 155/2 meters

Explanation:

Let X (t) represent the equation of the position, then you have to d2x / dt2 = 5.

Applying the fundamental theorem of the calculation dx/dt = 5t + vo. The speed equation is V (t) = 5t + vo. Since the initial velocity is 30m/s, V (0) = 5 (0) + vo = 30. Therefore, V (t) = dx/dt = 5t + 30. Applying again the fundamental theorem of the calculation X (t) = 5t^2 / 2 + 30t + xo.

Displacement in 5 seconds is given by X (5) - X (0).

X (5) - X (0) = 5 (5)^2/2 +3 (5) + Xo - 5 (0)^2/2 -3 (0) -Xo = 155/2

Displacement after 5 seconds is 155/2 meters

Brut [27]3 years ago

7 0

Answer:

Velocity at 5 seconds: 180 m/s

Ditance: 525 meters

Explanation:

In order to calculate displacement in this option we need to use the formula for distance when an object is experimenting aceleration:

$s=VoT+1/2at^2$

So now we just have to insert the data that we know:

$s=VoT+1/2at^2\\s= 30*5+ 1/2*30*25\\s=150+15*25\\s= 150+375\\s=525 m$

The distance after 5 seconds with an initial velocity of 30 m/s and an acceleration of 30 m/s will be 525 meters.

And the final velocity is calculated by the next formula:

$Vf= Vo+ A*t\\Vf= 30 m/s+ 30m/s*5\\Vf=180m/s$

So the final velocity at 5s would be 180 m/s

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