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2 years ago

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A jeep starts from rest with a constant acceleration of 4m/s2.At the same time a car travels with a constant speed of 36km/h ove

rtake and passes the jeep how far beyond the starting point will the jeep overtakes the car?

1 answer:

Phantasy [73]2 years ago

5 0

Answer:

25m

Explanation:

Let's assume the Jeep attains a velocity of 36km/h ; a constant speed same with that of the car.

While the Jeep is accelerating to that speed, the car with that speed passes it.

Now we can calculate the time taken for the Jeep to attain the velocity of 36km/h on her constant acceleration.

This time is t = v/a; from Newton's Law of Motion:

a = V-U / t ; a-acceleration

V is final velocity = 36km/h

U is initial velocity 0 since the body starts from rest.

Hence t = 36000/3600 ÷ 4 = 2.5s

Note conversting from km/h to m/s we multiply by 1000/3600.

But the distance covered by the car while the Jeep just accelerates is

S = U × t = 10× 2.5 = 25m.

Note From Newton's law of Motion, distance for constant speed is defined as: U × t

Hence the Car would be 25m off the starting point just as the Jeep accelerates. It would overtake the Jeep when it just covers 25m from the Jeep starting point.

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Answer:

Explanation:

height of pole = 15 ft

height of man = 6 ft

Let the length of shadow is y .

According to the diagram

Let at any time the distance of man is x.

The two triangles are similar

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15 y - 15 x = 6 y

9 y = 15 x

$y=\frac{5}{3}x$

Differentiate with respect to time.

$\frac{dy}{dt}=\frac{5}{3}\frac{dx}{dt}$

As given, dx/dt = 4 ft/s

$\frac{dy}{dt}=\frac{5}{3}\times 4$

$\frac{dy}{dt}=\frac{20}{3}$ ft/s

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Sam boards a ski lift, and rides up the mountain at 6 miles per hour. once at the top, sam immediately begins skiing down the mo

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Let us divide this problem into two parts:
1) Sam rides up the mountain.
2) Sam rides down the mountain.

1.
Since speed is distance over time, as:
$v = \frac{s}{t}$

Therefore, distance would be:
$s_{up} = v_{up} * t_{up}$

Where s = distance,
v = speed,
t = time.

In the problem, Sam's speed while riding up is v = 6 miles/hour = (6 * 1609.34 / 60) = 160.934 meters/second(in SI Units). Plug this value in the above equation, you would get:

$s_{up} = 160.934 * t_{up}$ --- (A)

2.
As Sam rides down the mountain, the speed given is:
$v_{down} = 54 miles/h$

Convert it in SI units; the speed would be in SI unit:
v = 54 miles/hour = (54 * 1609.34 / 60) = 1448.406 meters/second(in SI Units). Plug this value in the distance equation, you would get:
$s_{down} = 1448.406 * t_{down}$

Since the $s_{up} = s_{down}$ , therefore,

$160.934 * t_{up} = 1448.406 * t_{down}$

=> $t_{up} = 9 t_{down}$

Now the condition is that the whole trip, up and down, takes 40 minutes(2400seconds), it means:
$t_{up} + t_{down} = 2400$

Plug in the value of $t_{up}$ in the above equation, you would get:
$t_{down} = 240$

Therefore,
$s_{down} = 1448.406*240$
$s_{down} = 347617.44$ meters (in relation to seconds)

$s_{down} = 5793.624$ meters (in relation to hours)

Now the last step is to convert meters into miles, you would get:
$s_{down} = 5793.624/1609.34 = 3.6miles$

So the answer is 3.6miles.

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Answer:

Speed of the duck and bullet just after hit will be 2.4752 m/sec

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Velocity of the bullet $v_2=100m/sec$

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$1.01v=2.5$

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The angular velocity of the orbit about the sun is:

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PART A)

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now the acceleration is given by

$a_c = \omega^2 R$

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PART B)

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$F_c = ma_c$

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Part C)

now monster moves inside at radius R = 1 m

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