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Arturiano [62]
4 years ago
12

A person takes a trip, driving with a constant speed of 94.5 km/h except for a 22.0 min rest stop. If the person's average speed

is 65.1 km/h, how much time is spent on the trip and how far does the person travel?
Physics
1 answer:
qwelly [4]4 years ago
4 0
From the average speed you can fix an equation:

Average speed = distance / time

You know the average speed = 65.1 kg / h, then

65.1 = distance / total time,

where total time is the time traveling plus 22.0 minutes

Call t the time treavelling and pass 22 minutes to hours:

65.1 = distance / [t + 22/60] ==> distance = [t + 22/60]*65.1

 
From the constant speed, you can fix a second equation

Constant speed = distance / time traveling

94.5 = distance / t ==> distance = 94.5 * t

The distance is the same in both equations, then you have:

[t +22/60] * 65.1 = 94.5 t

Now you can solve for t.

65.1t + 22*65.1/60 = 94.5t

94.5t - 65.1t = 22*65.1/60

29.4t = 23.87

t = 23.87 / 29.4

t  = 0.812 hours

distance = 94.5 km/h * 0.812 h = 76.7 km

Answers: 1) 0.81 hours, 2) 76.7 km


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where g is the gravitational acceleration and h is the height of the ball at highest point. At that point, therefore, the potential energy is maximum, while the kinetic energy is zero, and so the velocity is also zero.

(b) 9.8 m/s upward

We can find the velocity of the ball 1 s before reaching its highest point by using the equation:

a=\frac{v-u}{t}

where

a = g = -9.8 m/s^2 is the acceleration due to gravity, which is negative since it points downward

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u is the initial velocity

t = 1 s is the time interval

Solving for u, we find

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(c) -9.8 m/s

The change in velocity during the 1-s interval is given by

\Delta v = v -u

where

v = 0 is the final velocity (at the highest point)

u = 9.8 m/s is the initial velocity

Substituting, we find

\Delta v = 0 - (+9.8 m/s)=-9.8 m/s

(d) 9.8 m/s downward

We can find the velocity of the ball 1 s after reaching its highest point by using again the equation:

a=\frac{v-u}{t}

where this time we have

a = g = -9.8 m/s^2 is the acceleration due to gravity, still negative

v  is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

t = 1 s is the time interval

Solving for v, we find

v = u+at = 0 +(-9.8 m/s^2)(1 s)= -9.8 m/s

and the negative sign means it points downward.

(e) -9.8 m/s

The change in velocity during the 1-s interval is given by

\Delta v = v -u

where here we have

v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

Substituting, we find

\Delta v = -9.8 m/s - 0=-9.8 m/s

(f) -19.6 m/s

The change in velocity during the overall 2-s interval is given by

\Delta v = v -u

where in this case we have:

v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = +9.8 m/s is the initial velocity (1 s before reaching the highest point)

Substituting, we find

\Delta v = -9.8 m/s - (+9.8 m/s)=-19.6 m/s

(g) -9.8 m/s^2

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F=mg

where

m is the mass of the ball

g = -9.8 m/s^2 is the acceleration due to gravity

According to Newton's second law, the resultant of the forces acting on the body is equal to the product of mass and acceleration (a), so

mg = ma

which means that the acceleration is

a= g = -9.8 m/s^2

and the negative sign means it points downward.

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(b)

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capacitance after, C' = k x C = 80 x 1.258 x 10^-12 = 100.64 x 10^-12 C

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Potential difference after immersion, V' = V / k = 255 / 80 = 3.1875 V

(c) initial energy,

U = \frac{q^{2}}{2C}

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