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topjm [15]
3 years ago
15

An object is released from rest at time t = 0 and falls through the air, which exerts a resistive force such that the accelerati

on a of the object is given by a = g – bV, where V is the object's speed and b is a constant. If limiting cases for large and small values of t are considered, which of the following is a possible expression for the speed of the object as an explicit function of time?
a. v = g(l - e-bt)/b.
b. v = (gebt)/b.
c. v = gt - bt2.
d. v = (g + a)t/b.
e. v = V0 + gt, v0.
Physics
1 answer:
lubasha [3.4K]3 years ago
6 0

Answer:

The correct option is

a. v = g (1-e^{-bt})/b

Explanation:

Time at which the object start fall t = 0

The acceleration a is given by a = g - bV

Where V = Speed of the object

Speed V² = u² + 2·a·h

However with the drag force the object will approach terminal velocity as t becomes progressively larger whereby v will stop increasing

Option a. is the only option that has  limiting value of v which is in the range of g as t increases ∴ option a. is the correct option.

v = g (1-e^{-bt})/b  as t increases (1-e^{-bt}) → 1 s and v→ g/b m/s

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(a) The height of the cliff will be 8.26 meters.

(b) The time would it take to reach the ground will be 0.717 sec.

<h3>What is velocity?</h3><h3 />

The change of displacement with respect to time is defined as the velocity. Velocity is a vector quantity. it is a time-based component.

(a) The height of the cliff will be 8.26 meters.

According to Newton's second equation of motion

\rm H =ut-\frac{1}{2} gt^2 \\\\ \rm H =8\times 2.35-\frac{1}{2} 9.81 (2.35)^2\\\\\rm H =8.16 \; m

Hence The height of the cliff will be 8.26 meters.

(b)The time would it take to reach the ground will be 0.717 sec.

We must have the final velocity to find the time so;

\rm v^2=u^2+2gh\\\\ \rm v^2=8^2+2\times 9.81 \times 8.6 \\\\ \rm v= \sqrt{8^2+2\times 9.81 \times 8.6}\\\\\rm v=15.03 \;m/sec

According to Newton's third equation of motion ;

\rm v=u-gt \\\\ \rm t=\frac{v-u}{g} \\\\ \rm t=\frac{15.03-8}{9.81} \\\\ \rm t=0.717 sec.

Hence the time would it take to reach the ground will be 0.717 sec.

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A police car is traveling north on a straight road at a constant 16.0 m/s. An SUV traveling north at 30.0 m/s passes the police
Nastasia [14]

Answer:

It will take 15.55s for the police car to pass the SUV

Explanation:

We first have to establish that both the police car and the SUV will travel the same distance in the same amount of time. The police car is moving at constant velocity and the SUV is experiencing a deceleration. Thus we will use two distance fromulas (for constant and accelerated motions) with the same variable for t and x:

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2. x=x_{0}+v_{0}t+\frac{at^{2}}{2}

Since both cars will travel the same distance x, we can equal both formulas and solve for t:

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We simplify the fraction present and rearrange for our formula so that it equals 0:

0.9\frac{m}{s^{2}} t^{2}-14\frac{m}{s}t=0 \\\\ t(0.9\frac{m}{s^{2}}t-14\frac{m}{s})=0

In the very last step we factored a common factor t. There is two possible solutions to the equation at t=0 and:

0.9\frac{m}{s^{2}}t-14\frac{m}{s}=0 \\\\  0.9\frac{m}{s^{2}}t =14\frac{m}{s} \\\\ t =\frac{14\frac{m}{s}}{0.9\frac{m}{s^{2}}}=15.56s

What this means is that during the displacement of the police car and SUV, there will be two moments in time where they will be next to each other; at t=0 s (when the SUV passed the police car) and t=15.56s(when the police car catches up to the SUV)

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