Ask question

Ask question

All categories

3 years ago

12

An engineer in a locomotive sees a car stuck on the track at a railroad crossing in front of the train. When the engineer first

sees the car, the locomotive is 300 m from the crossing and its speed is 18 m/s. If the engineer’s reaction time is 0.45 s, what should be the magnitude of the minimum deceleration to avoid an accident? Answer in units of m/s 2 .

2 answers:

nordsb [41]3 years ago

8 0

Answer:

a=0.555m/s^2

Explanation:

First we find the distance traveled from the moment the engineer reacts to the car, assuming uniform movement

X=VT

X=(18)(0.45)=8.1m

then we find the distance at which the deceleration begins, which is obtained by subtracting the total distance with the inner result

X=300-8.1=291.9

finally we use the equation for constant acceleration

Vf=0 final speed

Vo=18m/s= initial speed

X=291.9m

(Vf^2-Vo^2)/2X=a

(0-18^2)/(2*291.9)=a

a=0.555m/s^2

Leona [35]3 years ago

5 0

Answer:

- 0.56 m/s 2

Explanation:

To calculate the magnitude of deceleration to avoid the collision

distance = 300 m

initial speed = 18 m/s

reaction time ( t ) = 0.45 s

first calculate the distance traveled by the engineer/train during his reaction time

D₁ = initial speed * reaction time = 18 * 0.45 = 8.1 m

Then the remaining distance before impact between the car and Train would be

= distance - distance traveled during reaction time

= 300 m - 8.1 m = 291.9 m

from the equation of motion of objects

V² = U² + 2ax (equation 1)

To avoid impact the final velocity( v ) would be ( 0 )

u = initial speed ( 18 m/s )

a = acceleration/deceleration ( ? )

x = remaining distance before collision ( 291.9 m )

hence equation 1 becomes

a = $-\frac{18^{2} }{2*291.9}$ = - $\frac{324}{583.8}$ = - 0.56 m/s 2

note: the answer is in negative because the Locomotive is decelerating and not accelerating

You might be interested in

A video game includes an asteroid that is programmed to move in a straight line across a 17-inch monitor according to the equati

Ainat [17]

Answer:

The asteroid's acceleration at this point is $2.71\ m/s^2$

Explanation:

The equation that governs the trajectory of asteroid is given by :

$x=6.5t-2.3t^3$

The velocity of asteroid is given by :

$v=\dfrac{dx}{dt}\\\\v=\dfrac{d(6.5t-2.3t^3)}{dt}\\\\v=6.5-6.9t^2$

At some point during the trip across the screen, the asteroid is at rest. It means, v = 0

So,

$6.5-6.9t^2=0\\\\t=0.971\ s$

Acceleration,

$a=\dfrac{dv}{dt}\\\\a=\dfrac{d(6.5-6.9t^2)}{dt}\\\\a=-13.8t$

Put t = 0.971 s

$a=-13.8\times 0.197\\\\a=-2.71\ m/s^2$

So, the asteroid's acceleration at this point is $2.71\ m/s^2$ and it is decelerating.

6 0

3 years ago

A stuntwoman is going to attempt a jump across a canyon that is 77 m wide. The ramp on the far side of the canyon is 25 m lower

liq [111]

initial speed of the stuntman is given as

$v = 28 m/s$

angle of inclination is given as

$\theta = 15 degree$

now the components of the velocity is given as

$v_x = 28 cos15 = 27.04 m/s$

$v_y = 28 sin15 = 7.25 m/s$

here it is given that the ramp on the far side of the canyon is 25 m lower than the ramp from which she will leave.

So the displacement in vertical direction is given as

$\delta y = -25 m$

$\delta y = v_y * t + \frac{1}{2} at^2$

$-25 = 7.25 * t - \frac{1}{2}*9.8* t^2$

by solving above equation we have

$t = 3.12 s$

Now in the above interval of time the horizontal distance moved by it is given by

$d_x = v_x * t$

$d_x = 27.04 * 3.12 = 84.4 m$

since the canyon width is 77 m which is less than the horizontal distance covered by the stuntman so here we can say that stuntman will cross the canyon.

5 0

3 years ago

At a certain place, Earth's magnetic field has magnitude B =0.703 gauss and is inclined downward at an angle of 75.4° to the hor

Irina18 [472]

Answer:

The charge flows in coulombs is

$dq=1.843x10^{-5}C$

Explanation:

The current magnitude of current is given by the resistance and the induced Emf as:

$I=N*\frac{dF}{Rdt}$

$\frac{dq}{dt}=\frac{dF}{Rdt}=dq=N*\frac{dF}{R}$

$dq=\frac{N*\beta*A*(Cos(\alpha_f)-Cos(\alpha_i}{R}$

$N=1300$ , $\beta=0.703$ , $A=\pi*r^2=\pi*0.10^2=0.01\pi m^2$ , $R=99.4+202=301.4$ Ω

$\alpha_f=14.6$ , $\alpha_i=165.4$

Replacing :

$dq=\frac{1300*0.703x10^{-4}*0.01\pi*(0.9667-(-0.9667))}{202+99.4}$

$dq=1.843x10^{-5}C$

5 0

3 years ago

Global Precipitation Measurement (GPM) is a tool scientists use to forecast weather. Which statements describe GPM? Select three

lyudmila [28]

Answer:

B.It is a satellite that collects data about rain and snow

C.Its orbit covers 90 percent of Earth’s surface

F.The sensors measure microwaves

5 0

3 years ago

Learning Goal: To be able to calculate work done by a constant force directed at different angles relative to displacement

svlad2 [7]

Answer:

Woke done, W = 4156.92 Joules

Explanation:

The work done by the force can be calculated as :

$W=F\times s$

$W=Fs\ cos\theta$

$\theta$ is the angle between force and the displacement

It is assumed to find the work done for the given parameters i.e.

Force, F = 30 N

Distance travelled, s = 160 m

Angle between force and displacement, $\theta=30$

Work done is given by :

$W=Fs\ cos\theta$

$W=30\times 160\ cos(30)$

W = 4156.92 Joules

So, the work done by the object is 4156.92 Joules. Hence, this is the required solution.

5 0

3 years ago