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pshichka [43]
3 years ago
8

A car is traveling at 108 km/h, stuck behind a slower car. Finally the road is clear and the car pulls over to make a pass. The

driver stomps on the gas pedal and accelerates up to a speed of 135 km/h. If it took 3.5 s to reach this speed, what is the average acceleration of the car?
Physics
1 answer:
mezya [45]3 years ago
8 0

Answer:

The average acceleration of the car is 2.143 meters per square second.

Explanation:

Let assume that car accelerates uniformly, in that case, we can obtain the value of acceleration by using the following equation of motion:

v = v_{o}+a\cdot t

Where:

v_{o} - Initial velocity, measured in meters per second.

v - Final velocity, measured in meters per second.

a - Acceleration, measured in meters per square second.

t - Time, measured in seconds.

Now, we clear acceleration within expression:

a = \frac{v-v_{o}}{t}

Initial and final velocities are now converted from kilometers per hour into meters per second:

v_{o} = \left(108\,\frac{km}{h} \right)\cdot \left(1000\,\frac{m}{km} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}  \right)

v_{o} = 30\,\frac{m}{s}

v = \left(135\,\frac{km}{h} \right)\cdot \left(1000\,\frac{m}{km} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}  \right)

v = 37.5\,\frac{m}{s}

If we know that t = 3.5\,s, then, the average acceleration of the car is:

a = \frac{37.5\,\frac{m}{s}-30\,\frac{m}{s} }{3.5\,s}

a = 2.143\,\frac{m}{s^{2}}

The average acceleration of the car is 2.143 meters per square second.

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8 0
2 years ago
A bus leaves at 9 am with a group of tourists. They travel 350 km before they stop for lunch. Then they travel an additional 250
Anettt [7]

Average speed = Distance traveled / time taken

In this case Time taken = Difference in hours between 3 PM and 9 AM

                                        = 6 hours

Total distance traveled = 350 km + 250 km

                                       = 600 kilometers

So average speed = 600/6 = 100 km/hr

Average speed of bus = 27.78 m/s

So the bus's average speed = 27.78 m/s or 100 km/hr.

8 0
3 years ago
I need to find the current resistance and voltage for each in this complicated circuit plz help
konstantin123 [22]

Explanation:

The 11Ω, 22Ω, and 33Ω resistors are in parallel.  That combination is in series with the 4Ω and 10Ω resistors.

The net resistance is:

R = 4Ω + 10Ω + 1/(1/11Ω + 1/22Ω + 1/33Ω)

R = 20Ω

Using Ohm's law, we can find the current going through the 4Ω and 10Ω resistors:

V = IR

120 V = I (20Ω)

I = 6 A

So the voltage drops are:

V = (4Ω) (6A) = 24 V

V = (10Ω) (6A) = 60 V

That means the voltage drop across the 11Ω, 22Ω, and 33Ω resistors is:

V = 120 V − 24 V − 60 V

V = 36 V

So the currents are:

I = 36 V / 11 Ω = 3.27 A

I = 36 V / 22 Ω = 1.64 A

I = 36 V / 33 Ω = 1.09 A

If we wanted to, we could also show this using Kirchhoff's laws.

7 0
2 years ago
PLSLPSLLPSLPSLPS HELP I ONLY HAVE 5 MINS PLEASE
UkoKoshka [18]

Answer:

0.19m/s²

Explanation:

Initial velocity(u) = 50×1000/60×60

=13.88 m/s

Final velocity(v) = 36.5×1000/60×60

=10.13 m/s

Acceleration(a) = v-u/t

=10.13-13.88/19.5

a= -0.19m/s²

-a = 0.19m/s²

The magnitude of retar dation is 0.19m/s²

5 0
1 year ago
Read 2 more answers
A spinning wheel is slowed down by a brake, giving it a constant angular acceleration of ?5.20 rad/s2. during a 3.80-s time inte
ddd [48]

<span>We can answer this using the rotational version of the kinematic equations:</span><span>
θ = θ₀ + ω₀<span>t + ½αt²     -----> 1</span></span>

ω² = ω₀² + 2αθ            -----> 2

Where:

θ = final angular displacement = 70.4 rad

θ₀ = initial angular displacement = 0

ω₀ = initial angular speed

ω = final angular speed

t = time = 3.80 s

α = angular acceleration = -5.20 rad/s^2

Substituting the values into equation 1:<span>
70.4 = 0 + ω₀(3.80) + ½(-5.20)(3.80)² </span><span>

ω₀ = (70.4 + 37.544) / 3.80 </span><span>

ω₀ = 28.406 rad/s </span><span>


Using equation 2:
ω² = (28.406)² + 2(-5.2)70.4 


ω = 8.65 rad/s 


</span>

5 0
2 years ago
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