A car mass of 1330 kg is traveling at 28.0 m/s. The driver applies the brakes to bring the car to a rest over a distance of 79.0 m. Calculate the retarding force.

Vo = sqrt(2 a X)

Solve that for the acceleration, a.
X = the stopping distance, 79 m

a = Vo^2/(2X)= 4.96 m/s^2

Themn use F = M a

Two men pull a 31-kg box with forces 9.7 N and 7.6 N. Find the resultant scceleration of the box.

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Your question cannot be answered unless the directions of the two pulling forces are specified

does "sqrt" mean square root?

To calculate the retarding force, we can use the equation of motion:

\( F = \frac{{\Delta p}}{{\Delta t}} \)

Where:
- \( F \) is the force,
- \( \Delta p \) is the change in momentum,
- \( \Delta t \) is the change in time.

First, let's calculate the initial momentum of the car. Momentum is given by the equation:

\( p_i = m \cdot v \)

Where:
- \( p_i \) is the initial momentum,
- \( m \) is the mass of the car,
- \( v \) is the velocity of the car.

Substituting the given values:

\( p_i = 1330 \, \text{kg} \times 28.0 \, \text{m/s} \)

\( p_i = 37240 \, \text{kg} \cdot \text{m/s} \)

Next, we'll calculate the final momentum of the car, which is zero because the car comes to rest.

\( p_f = 0 \, \text{kg} \cdot \text{m/s} \)

Now, we can calculate the change in momentum:

\( \Delta p = p_f - p_i \)

\( \Delta p = 0 \, \text{kg} \cdot \text{m/s} - 37240 \, \text{kg} \cdot \text{m/s} \)

\( \Delta p = -37240 \, \text{kg} \cdot \text{m/s} \)

The negative sign indicates that the momentum has changed in the opposite direction.

Finally, we'll calculate the retarding force using the equation of motion:

\( F = \frac{{\Delta p}}{{\Delta t}} \)

The time it takes to stop can be calculated using the equation:

\( t = \frac{{\Delta x}}{{v}} \)

Where:
- \( t \) is the time it takes to stop,
- \( \Delta x \) is the distance the car traveled to stop,
- \( v \) is the initial velocity of the car.

Substituting the given values:

\( t = \frac{{79.0 \, \text{m}}}{{28.0 \, \text{m/s}}} \)

\( t \approx 2.821 \, \text{s} \)

Now, we can calculate the retarding force:

\( F = \frac{{-37240 \, \text{kg} \cdot \text{m/s}}}{{2.821 \, \text{s}}} \)

\( F \approx -13178 \, \text{N} \)

The retarding force is approximately -13178 N, where the negative sign indicates that the force is acting in the opposite direction to the initial motion of the car.