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VladimirAG [237]

3 years ago

7

A car is travelling at a speed of 30 m/s. It decelerates at a rate of 12 m/s^2. Calculate the time taken for the car to come to

a rest.

2 answers:

user100 [1]3 years ago

8 0

Answer:

The answer to your question is: t = 2.5 s

Explanation:

Data

vo = 30 m/s

a = -12 m/s2

t = ?

vf = 0 m/s

Formula

vf = vo + at

Substitution

0 m/s = 30 + (-12)t

Solve it for t -30 = -12t

t = -30 / -12

t = 30/12 = 15/6 = 5/2

t = 2.5 s

AveGali [126]3 years ago

6 0

Answer:

2.5 s.

Explanation:

The car is travelling initially with a speed of , $u=30m/s$

And the deacceleration of the car is, $a=-12m/s^{2}$

The car comes to rest so, final velocity will be zero.

From the first equation of motion.

$v=u+at$

Here, u is the initial velocity, v is the final velocity, a is the acceleration and t is the time taken.

Put all the variables from above.

$0=30+(-12)t\\t=\frac{30}{12}\\t=2.5 s$

Therefore, the time taken by the car comes to rest is 2.5 s.

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Answer:

Explanation:

Step one:

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final velocity v=?

Step two:

applying the first equation of motion

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3 years ago

Two hypothetical planets of masses m1 and m2 and radii r1 and r 2 , respectively, are nearly at rest when they are an infinite d

Leto [7]

Answer:

(a) $v_1 = m_2\sqrt{\frac{2G}{d(m_1+m_2)} }$

$v_2=m_1\sqrt{\frac{2G}{d(m_1+m_2)} }$

(b) Kinetic Energy of planet with mass m₁, is KE₁ = 1.068×10³² J

Kinetic Energy of planet with mass m₂, KE₂ = 2.6696×10³¹ J

Explanation:

Here we have when their distance is d apart

$F_{1} = F_{2} =G\frac{m_{1}m_{2}}{d^{2}}$

Energy is given by

$Energy \,of \,attraction = -G\frac{m_{1}m_{2}}{d}}+\frac{1}{2} m_{1} v^2_1+ \frac{1}{2} m_{2} v^2_2$

Conservation of linear momentum gives

m₁·v₁ = m₂·v₂

From which

v₂ = m₁·v₁/m₂

At equilibrium, we have;

$G\frac{m_{1}m_{2}}{d}} = \frac{1}{2} m_{1} v^2_1+ \frac{1}{2} m_{2} v^2_2$ which gives

$2G{m_{1}m_{2}}= d m_{1} v^2_1+ dm_{2} (\frac{m_1}{m_2}v_1)^2= dv^2_1(m_1+(\frac{m_1}{m_2} )^2)$

multiplying both sides by m₂/m₁, we have

$2Gm^2_{2}}= dv^2_1 m_2+dm_1v^2_1 =dv^2_1( m_2+m_1)$

Such that v₁ = $\sqrt{\frac{2Gm^2_2}{d(m_1+m_2)} }$

$v_1 = m_2\sqrt{\frac{2G}{d(m_1+m_2)} }$

Similarly, with v₁ = m₂·v₂/m₁, we have

$G\frac{m_{1}m_{2}}{d}} = \frac{1}{2} m_{1} v^2_1+ \frac{1}{2} m_{2} v^2_2\Rightarrow 2G{m_{1}m_{2}}= dm_{1} (\frac{m_2}{m_1}v_1)^2 +d m_{2} v^2_2= dv^2_2(m_2+(\frac{m_2}{m_1} )^2)$

From which we have;

$2G{m^2_{1}}= dm_{2} v_2^2 +d m_{1} v^2_2$ and

$v_2=m_1\sqrt{\frac{2G}{d(m_1+m_2)} }$

The relative velocity = v₁ + v₂ = $v_1+v_2=m_1\sqrt{\frac{2G}{d(m_1+m_2)} } + m_2\sqrt{\frac{2G}{d(m_1+m_2)} } = (m_1+m_2)\sqrt{\frac{2G}{d(m_1+m_2)} }$

v₁ + v₂ = $(m_1+m_2)\sqrt{\frac{2G}{d(m_1+m_2)} }$

(b) The kinetic energy KE = $\frac{1}{2}mv^2$

$KE_1= \frac{1}{2} m_{1} v^2_1 \, \, \, KE_2= \frac{1}{2} m_{2} v^2_2$

Just before they collide, d = r₁ + r₂ = 3×10⁶+5×10⁶ = 8×10⁶ m

$v_1 = 8\times10^{24}\sqrt{\frac{2\times6.67408 \times 10^{-11}} {8\times10^6(2.00\times10^{24}+8.00\times10^{24})} }$ = 10333.696 m/s

$v_2 = 2\times10^{24}\sqrt{\frac{2\times6.67408 \times 10^{-11}} {8\times10^6(2.00\times10^{24}+8.00\times10^{24})} }$ =2583.424 m/s

KE₁ = 0.5×2.0×10²⁴× 10333.696² = 1.068×10³² J

KE₂ = 0.5×8.0×10²⁴× 2583.424² = 2.6696×10³¹ J.

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