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andrezito [222]
3 years ago
6

A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find th

e velocity and the speed when t = 4. f(t) = 12 + 35 t + 1
Physics
1 answer:
Igoryamba3 years ago
5 0

A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find the velocity and the speed when t = 4. f(t) = 12t² + 35 t + 1

Answer:

Velocity = 131 m/s

Speed = 131 m/s

Explanation:

Equation of motion, s = f(t) = 12t² + 35 t + 1

To get velocity of the particle, let us find the first derivative of s

v (t) = ds/dt = 24t + 35

At t = 4

v(4) = 24(4) + 35

v(4) = 131 m/s

Speed is the magnitude of velocity. Since the velocity is already positive, speed is also 131 m/s

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Answer:

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Answer:

Proof in explanataion

Explanation:

The basic dimensions are as follows:

MASS = M

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i)

Given equation is:

H = \frac{u^2Sin^2\phi}{2g}

where,

H = height (meters)

u = speed (m/s)

g = acceleration due to gravity (m/s²)

Sin Ф = constant (no unit)

So there dimensions will be:

H = [L]

u = [LT⁻¹]

g = [LT⁻²]

Sin Ф = no dimension

Therefore,

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<u>[L] = [L]</u>

Hence, the equation is proven to be homogenous.

ii)

F = \frac{Gm_1m_2}{r^2}\\\\

where,

F = Force = Newton = kg.m/s² = [MLT⁻²]

G = Gravitational Constant = N.m²/kg² = (kg.m/s²)m²/kg² = m³/kg.s²

G = [M⁻¹L³T⁻²]

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Therefore,

[MLT^{-2}] = \frac{[M^{-1}L^{3}T^{-2}][M][M]}{[L]^2}\\\\\ [MLT^{-2}] = [M^{(-1+1+1)}L^{(3-2)}T^{-2}]\\\\

<u>[MLT⁻²] = [MLT⁻²]</u>

Hence, the equation is proven to be homogenous.

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