On its own, a certain tow-truck has a maximum acceleration of 3.0 m/s2. what would be the maximum acceleration when this truck was towing a bus of twice its own mass?

Let's call [tex]a_1=3 m/s^2[/tex] the maximum acceleration of the truck alone. The force produced by the engine to accelerate the truck in this situation is
[tex]F= ma_1[/tex]
with m being the mass of the truck alone.

If we attach a bus of twice the mass of the truck, now the whole system (truck+bus) has a mass of (m+2m)=3m. And in this case, the force produced by the engine is
[tex]F=3m a_2[/tex]
with a2 being the new acceleration.
The engine is always the same, so the force produced is still the same, so we can equalize F written in the first equation and in the second equation:
[tex]m a_1 = 3 m a_2[/tex]
and find a2, the new acceleration:
[tex]a_2 = \frac{m a_1}{3 m}= \frac{a_1}{3}= \frac{3 m/s^2}{3}=1 m/s^2 [/tex]

johanrusli johanrusli · Answer 2 · 2020-01-16T08:48:23+01:00

The maximum acceleration would be 1.0 m/s² when this truck was towing a bus of twice its own mass

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Further explanation

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

[tex]\boxed {F = ma }[/tex]

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

[tex]\texttt{ }[/tex]

Given:

initial acceleration = a₁ = 3.0 m/s²

mass of tow-truck = m₁ = m

mass of bus = m₂ = 2m

Asked:

final acceleration = a₂ = ?

Solution:

[tex]\texttt{Initial Force of Tow-Truck = Final Force of Tow-Truck }[/tex]

[tex]F_1 = F_2[/tex]

[tex]m_1 a_1 = m_2 a_2[/tex]

[tex]m (3.0) = (m + 2m) a_2[/tex]

[tex]3m = 3m (a_2)[/tex]

[tex]a_2 = 3m \div 3m[/tex]

[tex]\boxed{a_2 = 1.0 ~ \mathtt{ m/s^2}}[/tex]

[tex]\texttt{ }[/tex]

Conclusion :

The maximum acceleration would be 1.0 m/s²

[tex]\texttt{ }[/tex]

Learn more

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Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics