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3 years ago

At highway speeds a car can accelerate at 1.7 m/s2.

Physics

1 answer:

Vaselesa [24]3 years ago

6 0

Answer: 2.5 seconds

Explanation:

We know that the acceleration is:

a(t) = 1.7 m/s^2

To get the velocity function, we must integrate over time, and we will get:

v(t) = (1.7m/s^2)*t + v0

Where v0 is the initial velocity, in this case, we assume that we start at 23.6m/s, then the initial velocity is:

v0 = 23.6 m/s

Then the velocity equation is:

v(t) = (1.7m/s^2)*t + 23.6 m/s

Now we want to find the value of t such v(t) = 27.8 m/s

Then:

v(t) = 27.8 m/s = (1.7m/s^2)*t + 23.6 m/s

27.8 m/s - 23.6 m/s = (1.7m/s^2)*t

4.2 m/s = (1.7m/s^2)*t

4.2m/s/(1.7m/s^2) = t = 2.5 s

Then at that acceleration, you need 2.5 seconds.

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A block of mass M=10 kg is on a frictionless surface as shown in the photo attached. And it's attached to a wall by two springs

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i. the speed of the block of mass when the springs are connected in parallel is 7.07 A m/s

ii. the angular velocity when the two springs are in parallel is 7.07 rad/s

i. the speed of the block of mass when the springs are connected in series is 11.2 A m/s

ii. the angular velocity when the two springs are in series is 11.2 rad/s

<h3>a. </h3><h3>i. How to calculate the velocity of the mass when the springs are connected in parallel?</h3>

Since k is the spring constant of both springs = 250 N/m. The equivalent spring constant in parallel is k' = k + k

= 2k

= 2 × 250 N/m

= 500 N/m

Now since A is the maximum distance the block is pulled from its equilibrium position, the total energy of the block is E = 1/2kA

Also, 1/2k'A² = 1/2k'x² + 1/2Mv² where

k' = equivalent spring constant in parallel = 500 N/m,
A = maximum displacement of spring,
x = equilibrium position = 0 m,
M = mass of block = 10 kg and
v = speed of block at equilibrium position

Making v subject of the formula, we have

v = √[k'(A² - x²)/M]

Substituting the values of the variables into the equation, we have

v = √[k'(A² - x²)/M]

v = √[500 N/m(A² - (0)²)/10]

v = √[50 N/m(A² - 0)]

v = [√50]A m/s

v = [5√2] A m/s

v = 7.07 A m/s

So, the speed of the block of mass when the springs are connected in parallel is 7.07 A m/s

<h3>ii. The angular velocity of mass when the springs are in parallel</h3>

Since velocity of spring v = ω√(A² - x²) where

ω = angular velocity of spring,
A = maximum displacement of spring and
x = equilbrium position of spring = 0 m

Making ω subject of the formula, we have

ω = v/√(A² - x²)

Since v = 7.07 A m/s

Substituting the values of the other variables into the equation, we have

ω = v/√(A² - x²)

ω = 7.07 A m/s/√(A² - 0²)

ω = 7.07 A m/s/√(A² - 0)

ω = 7.07 A m/s/√A²

ω = 7.07 A m/s/A m

ω = 7.07 rad/s

So, the angular velocity when the two springs are in parallel is 7.07 rad/s

<h3>b. </h3><h3>i. How to calculate the velocity of the mass when the springs are connected in series?</h3>

Since k is the spring constant of both springs = 250 N/m. The equivalent spring constant in parallel is 1/k" = 1/k + 1/k

= 2/k

⇒ k" = k/2

k" = 250 N/m ÷ 2

= 125 N/m

Now since A is the maximum distance the block is pulled from its equilibrium position, the total energy of the block is E = 1/2kA

Also, 1/2k"A² = 1/2k"x² + 1/2Mv'² where

k" = equivalent spring constant in series = 125 N/m,
A = maximum displacement of spring,
x = equilibrium position = 0 m,
M = mass of block = 10 kg and
v' = speed of block at equilibrium position

Making v subject of the formula, we have

v = √[k"(A² - x²)/M]

Substituting the values of the variables into the equation, we have

v = √[k"(A² - x²)/M]

v = √[125 N/m(A² - (0)²)/10]

v = √[125 N/m(A² - 0)]

v = [√125]A m/s

v = [5√5] A m/s

v = 11.2 A m/s

So, the speed of the block of mass when the springs are connected in series is 11.2 A m/s

<h3>ii. The angular velocity of the mass when the springs are in series</h3>

Since velocity of spring v = ω√(A² - x²) where

ω = angular velocity of spring,
A = maximum displacement of spring and
x = equilbrium position of spring = 0 m

Making ω subject of the formula, we have

ω = v/√(A² - x²)

Since v = 11.2 A m/s

Substituting the values of the other variables into the equation, we have

ω = v/√(A² - x²)

ω = 11.2 A m/s/√(A² - 0²)

ω = 11.2 A m/s/√(A² - 0)

ω = 11.2 A m/s/√A²

ω = 11.2 A m/s/A m

ω = 11.2 rad/s

So, the angular velocity when the two springs are in series is 11.2 rad/s

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