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1 year ago

You drive 5 miles at 30 mi/hr and then another 3 miles at 50 mi/hr. What is your average speed for the whole8-mile trip?

Physics

1 answer:

s2008m [1.1K]1 year ago

6 0

Given data

The initial distance covered is d1 = 5 miles

The initial speed is s1 = 30 mi/hr

The final distance is d2 = 3 miles

The final speed is s2 = 50 mi/hr

The expression for the initial time taken to travel is given as:

$t_1=\frac{d_1}{s_1}$

The expression for the final time taken is given as:

$t_2=\frac{d_2}{s_2}$

The expression for the average speed for the whole trip is given as:

$\begin{gathered} \text{Average sp}eed\text{ =}\frac{total\text{ distance}}{total\text{ time}} \\ S_{Avg}=\frac{d_1+d_2}{t_1+t_2} \\ S_{Avg}=\frac{d_1+d_2}{\frac{d_1}{s_1}+\frac{d_2}{s_2}_{}} \end{gathered}$

Substitute the value in the above equation.

$\begin{gathered} S_{Avg}=\frac{5\text{ mi+3 mi}}{\frac{5\text{ mi}}{30\text{ mi/hr}}+\frac{3\text{ mi}}{50\text{ mi/hr}}} \\ S_{Avg}=35.3\text{ mi/hr} \end{gathered}$

Thus, the average speed for the whole trip is 35.3 mi/hr.

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Henry, whose mass is 95 kg, stands on a bathroom scale in an elevator. The scale reads 830 N for the first 3.8 s after the eleva

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v= 4.0 m/s

Explanation:

When standing on the bathroom scale within the moving elevator, there are two forces acting on Henry's mass: Normal force and gravity.
Gravity is always downward, and normal force is perpendicular to the surface on which the mass is located (the bathroom scale), in upward direction.
Normal force, can adopt any value needed to match the acceleration of the mass, according to Newton's 2nd Law.
Gravity (which we call weight near the Earth's surface) can be calculated as follows:

$F_{g} = m*g = 95 kg * 9.8 m/s2 = 930 N (1)$

According to Newton's 2nd Law, it must be met the following condition:

$F_{net} = F_{g} -F_{n} = m*a\\ F_{net} = 930 N - 830 N = 100 N = 95 Kg * a$

As the gravity is larger than normal force, this means that the acceleration is downward, so, we choose this direction as the positive.

Solving for a, we get:

$a =\frac{F_{net} }{m} =\frac{100 N}{95 kg} = 1.05 m/s2$

We can find the speed after the first 3.8 s (assuming a is constant), applying the definition of acceleration as the rate of change of velocity:

$v_{f} = a* t = 1.05 m/s * 3.8 m/s = 4.0 m/s$

Now, if during the next 3.8 s, normal force is 930 N (same as the weight), this means that both forces are equal each other, so net force is 0.
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As the elevator was moving, the only choice is that it is moving at a constant speed, the same that it had when the scale was read for the first time, i.e., 4 m/s downward.

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