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myrzilka [38]
3 years ago
6

The driver of a pickup truck going 89.9 km/h applies the brakes giving the truck a uniform deceleration of 5.99 m/s^2 while it t

ravels 19.3 m.
a. The velocity of the truck in m/s at the end of this displacement is
_________ m/s.
b. How much time has elapsed?
_________s
Physics
1 answer:
mel-nik [20]3 years ago
6 0

(a) First, convert the truck's velocity from km/h to m/s:

89.9\,\dfrac{\mathrm{km}}{\mathrm h}\cdot\dfrac{10^3\,\mathrm m}{\mathrm{km}}\cdot\dfrac{\mathrm h}{3600\,\mathrm s}=25.0\,\dfrac{\mathrm m}{\mathrm s}

Then the velocity v after traveling \Delta x=19.3\,\mathrm m satisfies

v^2-{v_0}^2=2a\Delta x

where v_0 and a are the truck's initial velocity and acceleration, respectively. So we have

v^2-\left(25.0\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-5.99\,\dfrac{\mathrm m}{\mathrm s^2}\right)(19.3\,\mathrm m)

\implies v^2=394\,\dfrac{\mathrm m^2}{\mathrm s^2}

\implies v=\pm19.8\,\dfrac{\mathrm m}{\mathrm s}

Taking the square root gives two possible velocities of the same magnitude but in opposite directions. It's technically feasible that over the given displacement, the truck changes its direction and is accelerating back towards its starting position since we're treating the truck as a particle.

(b) When acceleration is constant, average velocity is given by

v_{\mathrm{av}}=\dfrac{x-x_0}t=\dfrac{v+v_0}2

which tells us that

t=\dfrac{2(19.3\,\mathrm m)}{v+25.0\,\frac{\mathrm m}{\mathrm s}}

Plugging in both possible values of v we found earlier, we get t=0.862\,\mathrm s (if v is positive) or t=7.42\,\mathrm s (if v is negative).

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According to recent typical test data, a Ford Focus travels 0.240 mi in 19.3 s, starting from rest. The same car, when braking f
Anit [1.1K]

Answer:

Explanation:

a )

While breaking initial velocity u = 62.5 mph

= 62.5 x 1760 x 3 / (60 x 60 )  ft /s

= 91.66 ft / s

distance trvelled s = 150 ft

v² = u² - 2as

0 = 91.66²  - 2 a x 150

a = - 28 ft / s²

b ) While accelerating initial velocity u = 0

distance travelled s = .24 mi

time = 19.3 s

s = ut + 1/2 at²

s is distance travelled in time t with acceleration a ,

.24 = 0 + 1/2 a x 19.3²

a = .001288 mi/s²

= 2.06 m /s²

c )

If distance travelled s = .25 mi

final velocity v = ? a = .001288 mi / s²

v² = u² + 2as

= 0 + 2 x .001288 x .25

= .000644

v = .025 mi / s

= .0025 x 60 x 60 mi / h

= 91.35 mph .

d ) initial velocity u = 59 mph

= 86.53 ft / s

final velocity = 0

acceleration = - 28 ft /s²

v = u - at

0 = 86.53 - 28 t

t = 3 sec approx .

4 0
3 years ago
A ship sets out to sail to a point 120 km due north. an unexpected storm blows the ship to a point 100 km due east of its starti
Pavel [41]
If you draw the problem, it would look like that shown in the attached picture. The total length the ship will now travel can be solved using the Pythagorean theorem. The solution is as follows:

d = √(120)²+(100)²
d = 156.2 km

So, the ship will have to travel 156.2 km to the northwest direction.

8 0
3 years ago
If a car manufacturer reduced the mass of a new car model, so that it took a force of 4000 N
Sauron [17]

Answer:

1000 kg

Explanation:

Force = mass x acceleration, Hence, mass = Force/acceleration

mass = 4000/4 = 1000 kg

5 0
2 years ago
What will the distance from the point of origin if a car traveled at a speed of 25 km/hr for a time of 1 hour and 15 minutes?
liberstina [14]
Distance = velocity * time

1 hour and 15 minutes = 1.25 hours

25km/hr * 1.25 hours = 31.25km
6 0
3 years ago
For a given force as time ___
myrzilka [38]

Answer:

Increases

Explanation:

For a given force, as the time increases, the impulse related to that force increases also.

Impulse and time are directly related to one another. Impulse is mathematically given as:

           I  = F t

I is the impulse

F is the force

t is the time taken

So, as both force and time increases, the value of the impulse must increase.

8 0
2 years ago
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