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2 years ago

The speed of a car increases uniformly from 6 m/s to 42 m/s in 9 second calculate the acceleration of the car

Physics

1 answer:

djyliett [7]2 years ago

5 0

The acceleration of the car which changes the speed uniformly is 4 m/s²

<h3>What is acceleration?</h3>

Acceleration can be defined as the rate change of velocity with time.

acceleration a = (Δv) / (Δt)

Given is the initial velocity 6m/s and final velocity 42m/s, the time take fir this change in speed is 9s.

Substitute the values, we have

a = (42-6)/9

a = 36/9

a =4 m/s²

Thus, the acceleration is 4 m/s²

Learn more about acceleration.

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1 year ago

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3 years ago

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3 years ago

Read 2 more answers

A 0.600 kg block is attached to a spring with spring constant 15 N/m. While the block is sitting at rest, a student hits it with

gulaghasi [49]

Answer:

8.8 cm

31.422 cm/s

Explanation:

m = Mass of block = 0.6 kg

k = Spring constant = 15 N/m

x = Compression of spring

v = Velocity of block

A = Amplitude

As the energy of the system is conserved we have

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{0.6\times 0.44^2}{15}}\\\Rightarrow A=0.088\ m\\\Rightarrow A=8.8\ cm$

Amplitude of the oscillations is 8.8 cm

At x = 0.7 A

Again, as the energy of the system is conserved we have

$\dfrac{1}{2}kA^2=\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2\\\Rightarrow v=\sqrt{\dfrac{k(A^2-x^2)}{m}}\\\Rightarrow v=\sqrt{\dfrac{15(0.088^2-(0.7\times 0.088)^2)}{0.6}}\\\Rightarrow v=0.31422\ m/s$

The block's speed is 31.422 cm/s

4 0

3 years ago

The speed of a car increases uniformly from 6 m/s to 42 m/s in 9 second calculate the acceleration of the car​

The speed of a car increases uniformly from 6 m/s to 42 m/s in 9 second calculate the acceleration of the car