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Nikolay [14]
3 years ago
12

While strolling downtown on a Saturday afternoon, you stumble across an old car show. As you are walking along an alley toward a

main street, you glimpse a particularly stylish Alpha Romeo pass by. Tall buildings on either side of the alley obscure your view, so you see the car only as it passes between the buildings. Thinking back to your physics class, you realize that you can calculate the car's acceleration. You estimate the width of the alleyway between the two buildings to be 3 m. The car was in view for 0.4 s. You also heard the engine rev when the car started from a red light, so you know the Alpha Romeo started from rest 5 s before you first saw it. Find the magnitude of its acceleration.
Physics
1 answer:
snow_tiger [21]3 years ago
7 0

Answer:

1.44 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

v=u+at\\\Rightarrow v=0+a\times 5\\\Rightarrow v=5a

This velocity will be the initial velocity of the car when it passes through the first building

s=ut+\frac{1}{2}at^2\\\Rightarrow 3=5a\times 0.4+\frac{1}{2}\times a\times 0.4^2\\\Rightarrow a=1.44\ m/s^2

The acceleration of the car is 1.44 m/s²

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malfutka [58]

Answer:

Acceleration, a=-2.48\ m/s^2

Explanation:

Initial speed of the skater, u = 8.4 m/s

Final speed of the skater, v = 6.5 m/s

It hits a 5.7 m wide patch of rough ice, s = 5.7 m

We need to find the acceleration on the rough ice. The third equation of motion gives the relationship between the speed and the distance covered. Mathematically, it is given by :

v^2-u^2=2as

a=\dfrac{v^2-u^2}{2s}

a=\dfrac{(6.5)^2-(8.4)^2}{2\times 5.7}

a=-2.48\ m/s^2

So, the acceleration on the rough ice -2.48\ m/s^2 and negative sign shows deceleration.

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3 years ago
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Answer:

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72 km/h -72 km/h = 0 km/h

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