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2 years ago

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A bus travelling at 30m/s along a straight road passes throu a stationary taxi cab.the can begins to move 5seconds later within

a uniform acceleration of 2metres per seconds square in the same direction as the bus.1.calculate the time taken by the cab to meet the bus.2.distance the cab has travelled to catch-up with the bus

1 answer:

goldenfox [79]2 years ago

3 0

(1) The time taken by the cab to meet the bus is 18.96 seconds.

(2) The distance the cab has travelled to catch-up with the bus is 718.96 m.

<h3>Distance between the bus and the cab</h3>

The distance between the bus and the cab when the cab begins to move is calculated as follows;

D = speed x time

D = 30 m/s x 5 s

D = 150 m

<h3>Velocity of the cab</h3>

The velocity of the cab when it begins to move is calculated as follows;

v = at

(v - 30)t = D

(at - 30)t = 150

(2t - 30)t = 150

2t² - 30t = 150

2t² - 30t - 150 = 0

solve the quadratic equation using formula method

t = 18.96 seconds

<h3>Distance traveled by the cab</h3>

D = vt

D = 2t²

D = 2(18.96²)

D = 718.96 m

Thus, the time taken by the cab to meet the bus is 18.96 seconds. The distance the cab has travelled to catch-up with the bus is 718.96 m.

Learn more about distance here: brainly.com/question/2854969

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A cord of negligible mass runs around two massless, frictionless pulleys. A canister with mass m = 20 kg hangs from one pulley.

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(a) 196 N

The equation of the forces on the side of the cord where the force F is applied is:

$F-T=0$ (1)

where T is the tension in the cord.

On the other side of the cord, the equation of the forces on the canister is

$T-mg = ma$ (2)

where

m = 20 kg is the mass of the canister

$g=9.8 m/s^2$ is the acceleration of gravity

a is the acceleration

From (1),

$T=F$

Substituting into (2),

$F-mg = ma\\F=m(g+a)$

We want the canister to move at constant speed, so

a = 0

And therefore:

$F=mg=(20)(9.8)=196 N$

b) 2.0 cm

The cord is inextensible, this means that the acceleration of its parts are the same. Therefore, the acceleration of the free end must be the same as the acceleration of the canister: and this means that the two parts also cover the same distance in the same time.

Therefore, the free end of the cord must be moved exactly the same as the canister, by 2.0 cm.

c) 3.92 J, the same

The work done by the tension in the cord is

$W_T = T d$

where

T is the tension

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As we said in part (a), the tension in the cord is equal to the force applied to the free end:

T = F

So

T = 196 N

Therefore, the work done by the tension is

$W=(196)(0.02)=3.92 J$

And since the force applied (F) is the same, then the work done by you when pulling the cord is exactly the same.

(d) -3.92 J

The weight of the canister is

$F_g = mg =(20 kg)(9.8 m/s^2)=196 N$

However, the direction of the force of gravity is opposite to the displacement. Therefore, the work done by gravity is negative:

$W_g = - F_g d$

And substituting,

$W_g=-(196)(0.02)=-3.92 J$

(e) Zero

The net work done on the canister can be simply calculated by adding the work done by the tension in the cord and the weight of the canister:

$W=W_T+W_g = 3.92 + (-3.92 ) = 0$

This is in agreement with the work-energy theorem, which states that the work done on an object is equal to its change in kinetic energy. In this situation, the canister is moving at constant speed, so its kinetic energy is not change: therefore,

$\Delta K = 0$ (change in kinetic energy = 0)

and so, the work done on it is also zero.

(f) The pulley system changes the direction of the force applied

This is a simple pulley system, which means that the system does not multiply the force applied in input. In fact, the mechanical advantage of the system is

$MA=\frac{F_{out}}{F_{in}}$

where:

$F_{out}$ is the output force, which is the weight of the canister

$F_{in}$ is the force in input, which is F

So, the mechanical advantage is 1:

$MA=\frac{196 N}{196 N}=1$

From a point of view of energy, therefore, there is no advantage in this system.

However, the advantage offered by the pulley system concerns the direction of the force: in fact, it changes the direction of the applied force (which is F, downward) into the tension of the cord (which is upward on the canister).

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