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faltersainse [42]
3 years ago
7

A Honda Civic travels in a straight line along a road. The car’s distance x from a stop sign is given as a function of time t by

the equation x(t) = αt2 - βt3, where α = 1.50 m/s2 and β = 0.0500 m/s3. Calculate the average velocity of the car for each time interval: (a) t = 0 to t = 2.00 s; (b) t = 0 to t = 4.00 s; (c) t = 2.00 s to t = 4.00 s.
Physics
1 answer:
aleksklad [387]3 years ago
6 0

a) Average velocity: 2.8 m/s

b) Average velocity: 5.2 m/s

c) Average velocity: 7.6 m/s

Explanation:

a)

The position of the car as a function of time t is given by

x(t)=\alpha t^2 - \beta t^3

where

\alpha = 1.50 m/s^2

\beta = 0.05 m/s^3

The average velocity is given by the ratio between the displacement and the time taken:

v=\frac{\Delta x}{\Delta t}

The position at t = 0 is:

x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0

The position at t = 2.00 s is:

x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m

So the displacement is

\Delta x = x(2)-x(0)=5.6-0=5.6 m

The time interval is

\Delta t = 2.0 s - 0 s = 2.0 s

And so, the average velocity in this interval is

v=\frac{5.6 m}{2.0 s}=2.8 m/s

b)

The position at t = 0 is:

x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0

While the position at t = 4.00 s is:

x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m

So the displacement is

\Delta x = x(4)-x(0)=20.8-0=20.8 m

The time interval is

\Delta t = 4.0 - 0 = 4.0 s

So the average velocity here is

v=\frac{20.8}{4.0}=5.2 m/s

c)

The position at t = 2 s is:

x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m

While the position at t = 4 s is:

x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m

So the displacement is

\Delta x = 20.8 - 5.6 = 15.2 m

While the time interval is

\Delta t = 4.0 - 2.0 = 2.0 s

So the average velocity is

v=\frac{15.2}{2.0}=7.6 m/s

Learn more about average velocity:

brainly.com/question/8893949

brainly.com/question/5063905

#LearnwithBrainly

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Answer:

(a) E = 3.6 x 10³ N/C = 3.6 KN/C

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Explanation:

(a)

Electric field between oppositely charged plates is given as follows:

E = σ/ε₀

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E = Electric Field Intensity = ?

σ = surface charge density = 32 nC/m² = 3.2 x 10⁻⁸ C/m²

ε₀ = Permittivity of free space = 8.85 x 10⁻¹² C²/N.m²

Therefore,

E = (3.2 x 10⁻⁸ C/m²)/(8.85 x 10⁻¹² C²/N.m²)

<u>E = 3.6 x 10³ N/C = 3.6 KN/C</u>

<u></u>

(b)

E = ΔV/r

ΔV = Er

where,

r = distance between plates = 29 cm = 0.29 m

ΔV = Potential Difference = ?

ΔV = (3.6 x 10³)(0.29)

<u>V = 1044 Volts</u>

<u></u>

(c)

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K.E = F r

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K.E = E q r

where,

q = charge on proton = 1.6 x 10⁻¹⁹ C

Therefore,

K.E = (3600 N/C)(1.6 x 10⁻¹⁹ C)(0.29 m)

<u>K.E = 1.67 x 10⁻¹⁶ J</u>

<u></u>

(d)

K.E = (1/2)m(Vf² - Vi²)

where,

m = mass of proton = 1.67 x 10⁻²⁷ kg

Vf = Final Velocity = ?

Vi = Initial Velocity = 0 m/s

Therefore,

1.67 x 10⁻¹⁶ J = (1/2)(1.67 x 10⁻²⁷ kg)(Vf² - (0 m/s)²]

Vf² = (1.67 x 10⁻¹⁶ J)(2)/(1.67 x 10⁻²⁷ kg)

Vf = √(20 x 10¹⁰ m²/s²)

<u>Vf = 4.47 x 10⁵ m/s</u>

<u></u>

(e)

2as = Vf² - Vi²

2(a)(0.29 m) = (4.47 x 10⁵ m/s)² - (0 m/s)²

a = (20 x 10¹⁰ m²/s²)/0.58 m

<u>a = 3.45 x 10¹¹ m/s²</u>

<u></u>

(f)

F = ma

F = (1.67 x 10⁻²⁷ kg)(3.45 x 10¹¹ m/s²)

<u>F = 5.76 x 10⁻¹⁶ N</u>

<u></u>

(g)

E = F/q

E = (5.76 x 10⁻¹⁶ N)/(1.6 x 10⁻¹⁹ C)

<u>E = 3.6 x 10³ N/C = 3.6 KN/C</u>

<u></u>

(h)

<u>Both values are same in part (h) and (a)</u>

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