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3 years ago

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You are driving down the highway late one night at 20 m/s when a deer steps onto the road 35m in front of you. Your reaction tim

e before stepping on the brakes is 0.50s, and the maximum deceleration of your car is 10m/s^2. How much distance is between you and the deer when you come to a stop?Using the information given above and the value for distance that you found above, determine the time required for you to stop once you press the brakes.

1 answer:

8090 [49]3 years ago

6 0

Answer:

- Distance between car and the deer when the car stopped = 20 m

- The time required for you to stop once you press the brakes = less than 5 s in order not to hit the deer.

Explanation:

Using the equations of motion,

In the 0.5 s reaction time, we need to first calculate how far he has travelled in that time.

a = 0 m/s² (Since the car is travelling at constant velocity)

x = ?

Initial velocity = u = 20 m/s

x = ut + at²/2

x = 20×0.5 + 0 = 10 m

From that moment,

a = - 10 m/s²

u = initial velocity at the start of the deceleration = 10 m/s

v = final velocity = 0 m/s

x = ?

v² = u² + 2ax

0² = 10² + 2(-10)(x)

20x = 100

x = 5 m

Total distance travelled from when the deer stepped onto the road = 10 + 5 = 15 m

Distance between car and the deer when the car stopped = 35 - 15 = 20 m

b) To determine the time required to stop once you step on the brakes

u = 10 m/s

t = ?

v = 0 m/s²

x = distance from when the brake was stepped on to the deer = 35 - 10 = 25 m

x = (u + v)t/2

25 = (10 + 0)t/2

10t = 50

t = 5 s

Meaning the time required to stop once you step on the brakes is less than 5s.

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A pencil rolls horizontally of a 1 meter high desk and lands .25 meters from the base of the desk. How fast was the pencil rolli

vovikov84 [41]

Answer: 0.55 m/s

Explanation:

This situation is related to projectile motion (also called parabolic motion), where the main equations are as follows:

$x=V_{o} cos\theta t$ (1)

$y=y_{o}+Vo sin \theta t + \frac{g}{2}t^{2}$ (2)

Where:

$x=0.25 m$ is the horizontal displacement of the pencil

$V_{o}$ is the pencil's initial velocity

$\theta=0\°$ since we are told the pencil rolls <u>horizontally</u> before falling

$t$ is the time since the pencil falls until it hits the ground

$y_{o}=1 m$ is the initial height of the pencil

$y=0$ is the final height of the pencil (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity, always acting vertically downwards

Begining with (1):

$x=V_{o} cos(0\°) t$ (3)

$x=V_{o}t$ (4)

Finding $t$ from (2):

$0=1 m+ \frac{-9.8m/s^{2}}{2}t^{2}$ (5)

$t=\sqrt{\frac{-2y_{o}}{g}}$ (6)

Substituting (6) in (4):

$x=V_{o}\sqrt{\frac{-2y_{o}}{g}}$ (7)

Isolating $V_{o}$ :

$V_{o}=\frac{x}{\sqrt{\frac{-2y_{o}}{g}}}$ (8)

$V_{o}=\frac{0.25 m}{\sqrt{\frac{-2(1 m)}{-9.8m/s^{2}}}}$ (9)

Finally:

$V_{o}=0.55 m/s$

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3 years ago

Block 1, of mass m₁ = 1.30 kg , moves along a frictionless air track with speed v₁ = 29.0 m/s. It collides with block 2, of mass

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Answer:

a. 37.7 kgm/s b. 0.94 m/s c. -528.85 J

Explanation:

a. The initial momentum of block 1 of m₁ = 1.30 kg with speed v₁ = 29.0 m/s is p₁ = m₁v₁ = 1.30 kg × 29.0 m/s = 37.7 kgm/s

The initial momentum of block 2 of m₁ = 39.0 kg with speed v₂ = 0 m/s since it is initially at rest is p₁ = m₁v₁ = 39.0 kg × 0 m/s = 0 kgm/s

So, the magnitude of the total initial momentum of the two-block system = (37.7 + 0) kgm/s = 37.7 kgm/s

b. Since the blocks stick together after the collision, their final momentum is p₂ = (m₁ + m₂)v where v is the final speed of the two-block system.

p₂ = (1.3 + 39.0)v = 40.3v

From the principle of conservation of momentum,

p₁ = p₂

37.7 kgm/s = 40.3v

v = 37.7/40.3 = 0.94 m/s

So the final velocity of the two-block system is 0.94 m/s

c. The change in kinetic energy of the two-block system is ΔK = K₂ - K₁ where K₂ = final kinetic energy of the two-block system = 1/2(m₁ + m₂)v² and K₁ = final kinetic energy of the two-block system = 1/2m₁v₁²

So, ΔK = K₂ - K₁ = 1/2(m₁ + m₂)v² - 1/2m₁v₁² = 1/2(1.3 + 39.0) × 0.94² - 1/2 × 1.3 × 29.0² = 17.805 J - 546.65 J = -528.845 J ≅ -528.85 J

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A 10 kg block is attached to a light cord that is wrapped around the pulley of an electric motor, as shown above. If the motor r

Anika [276]

Answer:

156.8 Watts

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 10 kg

Height (h) = 8 m

Time (t) = 5 s

Power (P) =?

Next, we shall determine the energy used by the motor to raise the block. This can be obtained as follow:

Mass (m) = 10 kg

Height (h) = 8 m

Acceleration due to gravity (g) = 9.8 m/s²

Energy (E) =?

E = mgh

E = 10 × 9. 8 × 8

E = 784 J

Finally, we shall determine the power output of the motor. This can be obtained as illustrated below:

Time (t) = 5 s

Energy (E) = 784 J

Power (P) =?

P = E/t

P = 784 / 5

P = 156.8 Watts

Therefore, the power output of the motor is 156.8 Watts

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A 1500 kg car traveling at 15.0 m/s to the south collides with a 4500 kg truck that is at rest at a stopligt. The car comes to a

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Answer:

5 m/s, moving to the South.

Explanation:

Parameters given:

Mass of car, m = 1500 kg

Initial velocity of car, u = 15 m/s

Mass of truck, M = 4500 kg

Initial velocity of truck, v = 0 m/s (Truck is at rest)

Final velocity of car, U = 0 m/s (Car comes to a stop)

Final velocity of truck = V

Because the collision is elastic, we can apply the principle of conservation of momentum, we have that:

Total initial momentum = Total final momentum

m*u + M*U = m*v + M*V

(1500 * 15) + (4500 * 0) = (1500 * 0) + (4500 * V)

22500 + 0 = 0 + 00V

=> V = 22500/4500

V = 5 m/s

The velocity carries a positive sign, hence, it's moving in the same direction as the car was moving initially.

That is, it's moving to the South.

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