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Anestetic [448]
3 years ago
13

The velocity of a particle is given by v=20t² - 100t + 50, where v is in meters per second and t is in seconds. Evaluate the vel

ocity when a is zero and the distance travelled at that instant.
Physics
2 answers:
nadezda [96]3 years ago
7 0
<h2>Answer:</h2>

velocity =  50m/s

distance = -83.33m

<h2>Explanation:</h2>

The velocity of the particle is given by;

v = 20t² - 100t + 50        -------------------(i)

Since acceleration is the time rate of change in velocity, to get the acceleration (a), we find the derivative of equation (i) with respect to t as follows;

a = \frac{dv}{dt} = \frac{d(20t^{2}  - 100t + 50)}{dt}

a = 40t - 100            ------------------(ii)

Now, when a = 0, let's find the time t by substituting the value of a into equation (ii) as follows;

0 = 40t - 100

=> 40t = 100

=> t = \frac{100}{40}

=> t = 2.5seconds.

This means that at t = 2.5 seconds, the acceleration of the particle is zero(0)

(a) Now, to get the velocity at this instant (t = 2.5s), substitute the value of t into equation (i) as follows;

v = 20(0)² - 100(0) + 50

v = 0 - 0 + 50

v = 50 m/s

Therefore, the velocity when a is zero is 50m/s

(b) To get the distance (s) travelled at that instant, we integrate equation (i) as follows;

s = \int\limits^a_b {v} \, dt      -----------------------(iii)

Where;

a = the time instant = 2.5 seconds

b = the initial time instant = 0

v =  20t² - 100t + 50

Substitute these values into equation (iii) as follows;

s = \int\limits^a_b {(20t^{2} -100t + 50)} \, dt

s = \frac{20t^{3} }{3} - \frac{100t^{2} }{2} + 50t |^{a}_{b}

Substitute the values of a and b as follows;

s = [\frac{20(2.5)^{3} }{3} - \frac{100(2.5)^{2} }{2} + 50(2.5)]  - [\frac{20(0)^{3} }{3} - \frac{100(0)^{2} }{2} + 50(0)]

s =  [\frac{20(2.5)^{3} }{3} - \frac{100(2.5)^{2} }{2} + 50(2.5)] - 0

s = 104.17 - 312.5 + 125

s = -83.33m

Therefore, the distance traveled at that instant is -83.33m

allochka39001 [22]3 years ago
5 0

Explanation:

It is given that,

The velocity of a particle is given by :

v=20t^2-100t+50

Where

v is in m/s and t is in seconds

Let a is the acceleration of the object at time t. So,

a=\dfrac{dv}{dt}

a=\dfrac{d(20t^2-100t+50)}{dt}

a=40t-100

When a = 0

40t-100=0

t = 2.5 s

a is zero at t = 2.5 s. Velocity, v=20(2.5)^2-100(2.5)+50

v = -75 m/s

Since, v=\dfrac{ds}{dt}, s is the distance travelled

s=\int\limits{vdt}

s=\int\limits{(20t^2-100t+50)dt}

s=\dfrac{20t^3}{3}-50t^2+50t

At t = 2.5 s, s=\dfrac{20(2.5)^3}{3}-50(2.5)^2+50(2.5)

s = −83.34 m

Hence, this is the required solution.

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Answer:

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Explanation:

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Here since resistivity and length are constant, we only need to see how the resistance increases or decreases with change in area.

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Old Area = pi * D^2 / 4

The ratio of new area / old area is :

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Since area increases 9 times, and it is inversely proportional to resistance:

Resistance decreases by 9 times.

So, old resistance = Voltage / Current = 10 / 2 = 5 ohm

New Resistance = 5 / 9 = 0.5556 ohm    (decreases by 9 times)

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If a baseball has a zero velocity at some instant, is the acceleration of the baseball necessarily zero at that time? Explain -
ipn [44]

Answer:

No, not necessarily

Explanation:

If an object is moving with an acceleration that causes its speed to be reduced, there will be a moment in which it reaches v = 0, but this doesn't necessarily mean that the acceleration isn't acting anymore. If the object continues its movement with the same acceleration, it's velocity will become negative.

An example of an object that has zero velocity but non-zero acceleration:

If you throw an object in the air with a certain velocity, it will move vertically, reducing its velocity in a 9,8 m/s^{2} rate (which is the acceleration caused by gravity). At a certain point, the object will reach its maximum height, and will start to fall. In the exact moment that it reaches the maximum height, before it starts falling, its velocity is zero, but gravity is still acting on the object (this is the reason why it starts falling instead of just being stopped at that point). Therefore, at that point, the object has zero velocity but an acceleration of 9,8 m/s^{2}.

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3 years ago
A airplane travels 3260 kilometers in 4 hours. What is the airplane average speed?
zavuch27 [327]
I'm pretty sure that it's 815.
4 0
3 years ago
The question is in the picture
Sedbober [7]

Answer:

e) 120m/s

Explanation:

When the ball reaches its highest point, its velocity becomes zero, meaning

v_0-gt = 0.

where v_0 is the initial velocity.

Solving for t we get

t = \dfrac{v_0}{g}

which is the time it takes the ball to reach the highest point.

Now, after the ball has reached its highest point, it turns around and falls downwards. After time t_0 since it had reached the highest point, the ball has traveled downwards and the velocity v_f it has gained is

v_f = gt_0,

and we are told that this is twice the initial velocity v_0; therefore,

v_f = 2v_0  = gt_0

which gives

t_0 = \dfrac{2v_0}{g}.

Thus, the total time taken to reach velocity 2v_0 is

t_{tot} = t+t_0 = \dfrac{v_0}{g}+\dfrac{2v_0}{g}

t_{tot} = \dfrac{3v_0}{g}.

This t_{tot}, we are told, is 36 seconds; therefore,

36= \dfrac{3v_0}{g},

and solving for v_0 we get:

v_0 = \dfrac{36g}{3}

v_0 = \dfrac{36s(10m/s^2)}{3}

\boxed{v_0 = 120m/s}

which from the options given is choice e.

7 0
3 years ago
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