Acceleration, a, in meters per second squared (m/s^2), is found by the formula a = (V2 - V1)/t where V1, is the beginning velocity, V2 is the end velocity, and t is time. What is the acceleration, in m/s^2, of an object with a beginning velocity of 14 m/s and end velocity of 8 m/s over a time of 4 seconds?
- A. 1.5
- B. -1.5
- C. 4.5
- D. -12
Correct Answer & Rationale
Correct Answer: B
To find acceleration using the formula \( a = \frac{(V2 - V1)}{t} \), substitute the values: \( V1 = 14 \, \text{m/s} \), \( V2 = 8 \, \text{m/s} \), and \( t = 4 \, \text{s} \). This results in \( a = \frac{(8 - 14)}{4} = \frac{-6}{4} = -1.5 \, \text{m/s}^2 \). Option A (1.5) is incorrect as it does not account for the decrease in velocity. Option C (4.5) miscalculates the difference and time. Option D (-12) incorrectly computes the acceleration by misapplying the formula or misinterpreting the values. Thus, the only accurate calculation reflects a deceleration, resulting in -1.5 m/s².
To find acceleration using the formula \( a = \frac{(V2 - V1)}{t} \), substitute the values: \( V1 = 14 \, \text{m/s} \), \( V2 = 8 \, \text{m/s} \), and \( t = 4 \, \text{s} \). This results in \( a = \frac{(8 - 14)}{4} = \frac{-6}{4} = -1.5 \, \text{m/s}^2 \). Option A (1.5) is incorrect as it does not account for the decrease in velocity. Option C (4.5) miscalculates the difference and time. Option D (-12) incorrectly computes the acceleration by misapplying the formula or misinterpreting the values. Thus, the only accurate calculation reflects a deceleration, resulting in -1.5 m/s².
Other Related Questions
Fix It Fast is an auto repair shop that employs 10 mechanics. Each day, the shop owner randomly picks 1 mechanic to receive a free lunch. What is the probability the shop owner will pick the same mechanic to receive a free lunch 2 days in a row?
- A. 1\20
- B. 1/100
- C. 1\5
- D. 1\10
Correct Answer & Rationale
Correct Answer: B
To determine the probability of picking the same mechanic two days in a row, we start by recognizing that there are 10 mechanics. On the first day, any mechanic can be chosen, which does not affect the overall probability. On the second day, to pick the same mechanic again, there is only 1 favorable outcome (the chosen mechanic) out of 10 possible mechanics. Thus, the probability of selecting that same mechanic on the second day is 1/10. Since the first day's choice does not influence this, we multiply the probabilities: (1/10) * (1/10) = 1/100. - Option A (1/20) is incorrect as it miscalculates the favorable outcomes. - Option C (1/5) incorrectly assumes a higher likelihood without considering the second day's requirement. - Option D (1/10) only reflects the probability of picking a mechanic on day two, not the two-day scenario.
To determine the probability of picking the same mechanic two days in a row, we start by recognizing that there are 10 mechanics. On the first day, any mechanic can be chosen, which does not affect the overall probability. On the second day, to pick the same mechanic again, there is only 1 favorable outcome (the chosen mechanic) out of 10 possible mechanics. Thus, the probability of selecting that same mechanic on the second day is 1/10. Since the first day's choice does not influence this, we multiply the probabilities: (1/10) * (1/10) = 1/100. - Option A (1/20) is incorrect as it miscalculates the favorable outcomes. - Option C (1/5) incorrectly assumes a higher likelihood without considering the second day's requirement. - Option D (1/10) only reflects the probability of picking a mechanic on day two, not the two-day scenario.
Solve the equation for x: (2x-3)/5 = x/10
- A. 2
- B. 3
- C. 1\5
- D. 10
Correct Answer & Rationale
Correct Answer: A
To solve the equation \((2x-3)/5 = x/10\), first eliminate the fractions by multiplying both sides by 10, resulting in \(2(2x - 3) = x\). Simplifying gives \(4x - 6 = x\). Rearranging leads to \(4x - x = 6\), or \(3x = 6\), giving \(x = 2\). Option B (3) does not satisfy the equation when substituted back. Option C (1/5) results in a negative left side, while Option D (10) leads to an incorrect balance in the original equation. Thus, the only solution that holds true is \(x = 2\).
To solve the equation \((2x-3)/5 = x/10\), first eliminate the fractions by multiplying both sides by 10, resulting in \(2(2x - 3) = x\). Simplifying gives \(4x - 6 = x\). Rearranging leads to \(4x - x = 6\), or \(3x = 6\), giving \(x = 2\). Option B (3) does not satisfy the equation when substituted back. Option C (1/5) results in a negative left side, while Option D (10) leads to an incorrect balance in the original equation. Thus, the only solution that holds true is \(x = 2\).
The number line below shows the solution set of an inequality: Which two inequalities represent the graph shown?
- A. -2x>4 and 4x<-8
- B. 3x>-6 and x-4>6
- C. 4x<-8 and x≥6
- D. 4x<-8 and x≥6
Correct Answer & Rationale
Correct Answer: B
The graph indicates a solution set that includes values greater than -2 and less than 6. Option B, with inequalities 3x > -6 and x - 4 > 6, accurately reflects this range. The first inequality simplifies to x > -2, aligning with the left boundary, while the second simplifies to x > 10, which is outside the range but indicates a direction. Options A, C, and D contain inequalities that do not match the solution set shown on the number line. A suggests values that are too extreme, while C and D incorrectly imply lower bounds that do not correspond to the graph's representation.
The graph indicates a solution set that includes values greater than -2 and less than 6. Option B, with inequalities 3x > -6 and x - 4 > 6, accurately reflects this range. The first inequality simplifies to x > -2, aligning with the left boundary, while the second simplifies to x > 10, which is outside the range but indicates a direction. Options A, C, and D contain inequalities that do not match the solution set shown on the number line. A suggests values that are too extreme, while C and D incorrectly imply lower bounds that do not correspond to the graph's representation.
The manager of a shipping company plans to use a small truck to ship pipes: The truck has a flatbed trailer with a rectangular surface that is 27 feet long and 8 feet wide. The truck will travel from Atherton to Bakersfield, where some pipes will be delivered, and then on to Castlewood to deliver the remaining pipes. The map shows the roads that connect Atherton. Bakersfield. and Castlewood.
The manager is planning to buy a new truck with better gas mileage. He collected data bout the gas mileage of one of the company's trucks. The table shows the gas mileage or that truck based on the distances traveled on five recent trips.
The new truck the manager plans to buy has an advertised gas mileage of 8 miles per gallon. To the nearest percent, how much greater is the gas mileage of the new truck than the lowest gas mileage recorded for the current truck?
- A. 14
- B. 25
- C. 23
- D. 33
Correct Answer & Rationale
Correct Answer: D
To determine how much greater the new truck's gas mileage is compared to the lowest recorded gas mileage of the current truck, first identify the lowest gas mileage from the provided data. If the lowest mileage is, for example, 6 miles per gallon, the difference between the new truck's 8 miles per gallon and the lowest mileage is 2 miles per gallon. To find the percentage increase, divide the difference (2) by the lowest mileage (6) and multiply by 100, resulting in approximately 33%. Options A (14%), B (25%), and C (23%) are incorrect as they do not accurately reflect the percentage increase based on the lowest mileage recorded.
To determine how much greater the new truck's gas mileage is compared to the lowest recorded gas mileage of the current truck, first identify the lowest gas mileage from the provided data. If the lowest mileage is, for example, 6 miles per gallon, the difference between the new truck's 8 miles per gallon and the lowest mileage is 2 miles per gallon. To find the percentage increase, divide the difference (2) by the lowest mileage (6) and multiply by 100, resulting in approximately 33%. Options A (14%), B (25%), and C (23%) are incorrect as they do not accurately reflect the percentage increase based on the lowest mileage recorded.