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2 years ago

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Jade and her roommate Jari commute to work each morning, traveling west on I-10. One morning Jade left for work at 6:45 A.M., bu

t Jari left 15 minutes later. Both drove at a constant speed. The following graphs show the distance (in miles) each of them has traveled on I-10 at time t (in minutes), where t = 0 is 7:00 A.M.

1 answer:

Veronika [31]2 years ago

3 0

Answer:

Jari

Explanation:

The question requires to know who is traveling faster. This is done by comparing the gradients. The steeper the slope (high gradient), the faster the speed and vice versa.

From Jari's line, the starting point is (0, 0) and another point is (6, 7)

The gradient being change in y to change in x

Change in y=7-0=7

Change in x=6-0=6

Slope is 7/6

For Jade, first point is (0, 10) then another point is (6, 16)

Change in y=16-10=6

Change in x=6-0=6

Slope is 6/6=1

Clearly, 7/6 is greater than 6/6 or 1 hence Jari is faster than Jade

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How long does it take an automobile traveling 66.7 km/h to become even with a car that is traveling in another lane at 52.7 km/h

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Answer:

The time taken is $t = 32.5 \ s$

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From the question we are told that

The speed of first car is $v_1 = 66.7 \ km/h = 18.3 \ m/s$

The speed of second car is $v_2 = 52.7 \ km/h = 14.64 \ m/s$

The initial distance of separation is $d = 119 \ m$

The distance covered by first car is mathematically represented as

$d_t = d_i + d_f$

Here $d_i$ is the initial distance which is 0 m/s

and $d_f$ is the final distance covered which is evaluated as $d_f = v_1 * t$

So

$d_t = 0 \ m/s + (v_1 * t )$

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The distance covered by second car is mathematically represented as

$d_t = d_i + d_f$

Here $d_i$ is the initial distance which is 119 m

and $d_f$ is the final distance covered which is evaluated as $d_f = v_2* t$

$d_t = 119 + 14.64 * t$

Given that the two car are now in the same position we have that

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<em><u>or</u></em>

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3 years ago

For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is

ArbitrLikvidat [17]

Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀ $e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)$

where A(t)=A₀ $e^{\frac{-bt}{2m}}$

A₀ is the amplitude at t=0 and

$w'$ is the angular frequency of damped SHM, which is given by,

$w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

Now coming to the problem,

Given: m=1.2 kg

k=9.8 N/m

b=210 g/s= 0.21 kg/s

A₀=13 cm

a) A(t)=A₀/8

⇒A₀ $e^{\frac{-bt}{2m}}$ =A₀/8

⇒ $e^{\frac{bt}{2m}}=8$

applying logarithm on both sides

⇒ $\frac{bt}{2m}=ln(8)$

⇒ $t=\frac{2m*ln(8)}{b}$

substituting the values

$t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)$

b) $w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

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