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Alex Ar [27]
3 years ago
13

A car accelerates to the right from 12 m/s to 30 m/s in 6.0 s. What was its acceleration and displacement during this time inter

val
Physics
1 answer:
Scorpion4ik [409]3 years ago
7 0

Answer:

Acceleration = 3m/s^2

Displacement = 126 m

Explanation:

The velocity changed from 12 to 30 m/s in 6 seconds. Using the formula for acceleration we get

(30-12)/6= 18/6= 3m/s^2

Displacement is d=.5a*t^2+v0*t

Inputting acceleration, time, and velocity leaves us with

d=1.5 * 36 + 12 * 6 = 54 + 72 = 126 m

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A 4500 kg car accelerates from rest to 45.0
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The car undergoes an acceleration <em>a</em> such that

(45.0 km/h)² - 0² = 2 <em>a</em> (90 m)

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Solve for <em>a</em> :

<em>a</em> = (45.0 km/h)² / (2 (0.09 km)) = 11,250 km/h²

Ignoring friction, the net force acting on the car points in the direction of its movement (it's also pulled down by gravity, but the ground pushes back up). Newton's second law then says that the net force <em>F</em> is equal to the mass <em>m</em> times the acceleration <em>a</em>, so that

<em>F</em> = (4500 kg) (11,250 km/h²)

Recall that Newtons (N) are measured as

1 N = 1 kg • m/s²

so we should convert everything accordingly:

11,250 km/h² = (11,250 km/h²) (1000 m/km) (1/3600 h/s)² ≈ 0.868 m/s²

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8 0
4 years ago
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