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3 years ago

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A train starts from rest and accelerates uniformly until it has traveled 5.6 km and acquired a forward velocity of The train the

n moves at a constant velocity of for 420 s. The train then slows down uniformly at until it is brought to a halt. The acceleration during the first 5.6 km of travel is closest to which of the following?
a. 0.19 m/s^2
b. 0.14 m/s^2
c. 0.16 m/s^2
d. 0.20 m/s^2
e. 0.17 m/s^2

1 answer:

Korvikt [17]3 years ago

7 0

Answer:

The acceleration during the first 5.6 km of travel is closest to 0.16 m/s²

Option c) 0.16 m/s² is the correct answer.

Explanation:

Given the data in the question;

since the train starts from rest,

Initial velocity; u = 0 m/s

final velocity; v = 42 m/s

distance covered S = 5.6 km = ( 5.6 × 1000 )m = 5600 m

acceleration a = ?

From the third equation of motion;

v² = u² + 2as

we substitute in our values

( 42 )² = ( 0 )² + [ 2 × a × 5600 ]

1764 = 0 + [ 11200 × a ]

1764 = 11200 × a

a = 1764 / 11200

a = 0.1575 ≈ 0.16 m/s² { two decimal place }

Therefore, The acceleration during the first 5.6 km of travel is closest to 0.16 m/s²

Option c) 0.16 m/s² is the correct answer.

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a) Time at which velocity is +20.0 m/s: 2.04 s

b) Time at which velocity is -20.0 m/s: 6.12 s

c) Time at which the displacement is zero: t = 0 and t = 8.16 s

d) Time at which the velocity is zero: t = 4.08 s

e) i) ii) iii) The acceleration of the boulder is always $9.8 m/s^2$ downward

f) See graphs in attachment

Explanation:

a)

The motion of the boulder is a uniformly accelerated motion, with constant acceleration

$a=g=-9.8 m/s^2$

downward (acceleration due to gravity). So, we can use the following suvat equation:

$v=u+at$

where:

v is the velocity at time t

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a=g=-9.8 m/s^2 is the acceleration

We want to find the time t at which the velocity is

v = 20.0 m/s

Therefore,

$t=\frac{v-u}{a}=\frac{20-40}{-9.8}=2.04 s$

b)

In this case, we want to find the time t at which the boulder is moving at 20.0 m/s downward, so when

v = -20.0 m/s

(the negative sign means downward)

We use again the suvat equation

$v=u+at$

And substituting

u = +40.0 m/s

a=g=-9.8 m/s^2

We find the corresponding time t:

$t=\frac{v-u}{a}=\frac{-20-(+40)}{-9.8}=6.12 s$

c)

To solve this part, we can use the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where

s is the displacement

u = +40.0 m/s is the initial velocity

$a=g=-9.8 m/s^2$ is the acceleration

t is the time

We want to find the time t at which the displacement is zero, so when

s = 0

SUbstituting into the equation and solving for t,

$0=ut+\frac{1}{2}at^2\\t(u+\frac{1}{2}a)=0$

which gives two solutions:

t = 0 (initial instant)

$u+\frac{1}{2}at=0\\t=-\frac{2u}{a}=-\frac{2(40)}{-9.8}=8.16 s$

which is the instant at which the boulder passes again through the initial position, but moving downward.

d)

To solve this part, we can use again the suvat equation

$v=u+at$

where

u = +40.0 m/s is the initial velocity

$a=g=-9.8 m/s^2$ is the acceleration

We want to find the time t at which the velocity is zero, so when

v = 0

Substituting and solving for t, we find:

$t=\frac{v-u}{a}=\frac{0-(40)}{-9.8}=4.08 s$

e)

In order to evaluate the acceleration of the boulder, let's consider the forces acting on it.

If we neglect air resistance, there is only one force acting on the boulder: the force of gravity, acting downward, with magnitude

$F=mg$

where m is the mass of the boulder and $g$ the acceleration of gravity.

According to Newton's second law, the net force on the boulder is equal to the product between its mass and its acceleration:

$F=ma$

Combining the two equations, we get

$ma=mg\\a=g$

So, the acceleration of the boulder is $g=9.8 m/s^2$ downward at any point of the motion, no matter where the boulder is (because the force of gravity is constant during the motion).

f)

Find the three graphs in attachment:

- Position-time graph: the position of the boulder initially increases as the boulder goes upward; however, the slope of the curve decreases as the boulder goes higher (because the velocity decreases). The boulder reaches its maximum height at t = 4.08 s (when velocity is zero), then it starts going downward, until reaching its initial position at t = 8.16 s

- Velocity-time graph: the initial velocity is +40 m/s; then it decreases linearly (because the acceleration is constant), and becomes zero when t = 4.08 s. Then the velocity becomes negative (because the boulder is now moving downward) and its magnitude increases.

- Acceleration-time graph: the acceleration is constant and it is $-9.8 m/s^2$ , so this graph is a straight horizontal line.

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3 years ago