Questions: Lewis is towing a 1140 kg car. His truck pulls with a force of 2850 N and increases the car's velocity from 0.9 s to 8.1 s.

Solution Steps

We are given the following information:

• Mass of the car, \( m = 1140 \, \text{kg} \)
• Force applied by the truck, \( F = 2850 \, \text{N} \)
• Initial time, \( t_1 = 0.9 \, \text{s} \)
• Final time, \( t_2 = 8.1 \, \text{s} \)

Using Newton's second law, the acceleration \( a \) can be calculated as:

\[ a = \frac{F}{m} = \frac{2850 \, \text{N}}{1140 \, \text{kg}} = 2.5 \, \text{m/s}^2 \]

The change in velocity \( \Delta v \) can be calculated using the formula for acceleration:

\[ a = \frac{\Delta v}{\Delta t} \]

where \( \Delta t = t_2 - t_1 = 8.1 \, \text{s} - 0.9 \, \text{s} = 7.2 \, \text{s} \).

Rearranging the formula to solve for \( \Delta v \):

\[ \Delta v = a \cdot \Delta t = 2.5 \, \text{m/s}^2 \times 7.2 \, \text{s} = 18 \, \text{m/s} \]

Final Answer