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On its own, a tow truck has an acceleration of 3.0 m/s². What would be the acceleration when the truck is towing a bus twice its own mass? 2.0 m/s^2 O 1.5 m/s^2 O 1.0 m/s^2 2.5 m/s^2 O none of these

On its own, a tow truck has an acceleration of 3.0 m/s². What would be the acceleration when the truck is towing a bus twice its own mass? 2.0 m/s^2 O 1.5 m/s^2 O 1.0 m/s^2 2.5 m/s^2 O none of these

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**Physics Problem: Truck Acceleration** *Scenario:* A tow truck, by itself, can accelerate at a rate of \(3.0 \, \text{m/s}^2\). Consider the case where this truck is towing a bus that has twice the truck's own mass. What will be the new acceleration of the tow truck? *Options:* - \(2.0 \, \text{m/s}^2\) - \(1.5 \, \text{m/s}^2\) - \(1.0 \, \text{m/s}^2\) - \(2.5 \, \text{m/s}^2\) - none of these *Explanation:* To solve this problem, use Newton’s Second Law. The force exerted by the truck remains constant. Originally, the force \( F \) can be expressed as: \[ F = m \times a = m \times 3.0 \, \text{m/s}^2 \] When towing the bus, the total mass becomes \( 3m \), where \( m \) is the mass of the truck. With the same force applied: \[ F = (3m) \times a' \] Since the force \( F \) remains unchanged, equate the two expressions and solve for \( a' \) (new acceleration): \[ m \times 3.0 = 3m \times a' \] \[ a' = \frac{3.0}{3} = 1.0 \, \text{m/s}^2 \] Thus, the correct choice is \(1.0 \, \text{m/s}^2\).
Transcribed Image Text:**Physics Problem: Truck Acceleration** *Scenario:* A tow truck, by itself, can accelerate at a rate of \(3.0 \, \text{m/s}^2\). Consider the case where this truck is towing a bus that has twice the truck's own mass. What will be the new acceleration of the tow truck? *Options:* - \(2.0 \, \text{m/s}^2\) - \(1.5 \, \text{m/s}^2\) - \(1.0 \, \text{m/s}^2\) - \(2.5 \, \text{m/s}^2\) - none of these *Explanation:* To solve this problem, use Newton’s Second Law. The force exerted by the truck remains constant. Originally, the force \( F \) can be expressed as: \[ F = m \times a = m \times 3.0 \, \text{m/s}^2 \] When towing the bus, the total mass becomes \( 3m \), where \( m \) is the mass of the truck. With the same force applied: \[ F = (3m) \times a' \] Since the force \( F \) remains unchanged, equate the two expressions and solve for \( a' \) (new acceleration): \[ m \times 3.0 = 3m \times a' \] \[ a' = \frac{3.0}{3} = 1.0 \, \text{m/s}^2 \] Thus, the correct choice is \(1.0 \, \text{m/s}^2\).
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