Problem 20 - Applied Physics

Question

(a) What force is required to slow a \(1350-\mathrm{kg}\) car traveling \(90.0 \mathrm{~km} / \mathrm{h}\) to \(25.0 \mathrm{~km} / \mathrm{h}\) within \(4.00 \mathrm{~s}\) ? (b) How far does the car travel during its deceleration? (c) How long does it take for the car to come to a complete stop at this same rate of deceleration?

Step-by-Step Solution

Answer

(a) -6095.25 N, (b) 63.88 m, (c) 5.54 s.

Step 1: Convert Speeds to Meters per Second

First, convert the given speeds from kilometers per hour to meters per second. Use the conversion factor: \( 1 \, \text{km/h} = \frac{1}{3.6} \, \text{m/s}\).- Initial speed: \( 90.0 \, \text{km/h} = 90.0 \, \text{km/h} \times \frac{1}{3.6} = 25.0 \, \text{m/s}\)- Final speed: \( 25.0 \, \text{km/h} = 25.0 \, \text{km/h} \times \frac{1}{3.6} = 6.94 \, \text{m/s}\)

Step 2: Calculate Constant Acceleration

Use the formula for acceleration: \( a = \frac{v_f - v_i}{t} \), where \( v_f \) is the final speed, \( v_i \) is the initial speed, and \( t \) is the time.\[ a = \frac{6.94 \, \text{m/s} - 25.0 \, \text{m/s}}{4.00 \, \text{s}} = \frac{-18.06}{4.00} = -4.515 \, \text{m/s}^2 \]

Step 3: Calculate the Required Force

Use Newton's second law, \( F = ma \), where \( m \) is the mass of the car and \( a \) is the acceleration.\[ F = 1350 \, \text{kg} \times (-4.515 \, \text{m/s}^2) = -6095.25 \, \text{N} \]The negative sign indicates the force is in the opposite direction of motion.

Step 4: Calculate the Distance Traveled During Deceleration

Use the kinematic equation for distance: \( d = v_i t + \frac{1}{2} a t^2 \).\[ d = 25.0 \, \text{m/s} \times 4.00 \, \text{s} + \frac{1}{2} (-4.515 \, \text{m/s}^2) \times (4.00 \, \text{s})^2 \]\[ d = 100.0 \, \text{m} - 36.12 \, \text{m} = 63.88 \, \text{m} \]

Step 5: Time to Stop Completely

Use the formula for time with kinematic equation \( v_f = v_i + at \) and set \( v_f = 0 \).\[ 0 = 25.0 \, \text{m/s} + (-4.515 \, \text{m/s}^2) \times t \]\[ t = \frac{-25.0 \, \text{m/s}}{-4.515 \, \text{m/s}^2} = 5.54 \, \text{s} \]

Key Concepts

Acceleration

Acceleration is a measure of how quickly an object's velocity changes over time. It is a vector quantity, which means it has both a magnitude and a direction.In the exercise, we calculated the car's acceleration as it slows down from 90 km/h to 25 km/h.The formula we used was:\[a = \frac{v_f - v_i}{t}\]where:- \(a\) is the acceleration,- \(v_f\) is the final velocity,- \(v_i\) is the initial velocity,- \(t\) is the time.In our problem, we found an acceleration of \(-4.515 \, \text{m/s}^2\).The negative sign indicates the car is decelerating, or slowing down.Understanding acceleration helps us predict how an object's motion changes over time.

Newton's second law

Newton's second law of motion is crucial for understanding how forces affect the motion of objects.This law states:\[F = ma\]where:- \(F\) is the force,- \(m\) is the mass of the object, - \(a\) is the acceleration.In our exercise, the car's mass was 1350 kg, and the calculated acceleration was \(-4.515 \, \text{m/s}^2\).By applying Newton's second law, we found the required force to decelerate the car:\[F = 1350 \, \text{kg} \times (-4.515 \, \text{m/s}^2) = -6095.25 \, \text{N}\]This negative force indicates it's in the opposite direction to the car’s motion.Newton's second law helps us quantify the influence of forces on an object's motion.

Kinematic equations

Kinematic equations describe the motion of objects without considering the forces that cause the motion.These equations use variables like velocity, acceleration, and time to find other unknowns like distance or final velocity.In our exercise, we used a kinematic equation to find the distance:\[d = v_i t + \frac{1}{2} a t^2\]where:- \(d\) is the distance,- \(v_i\) is the initial velocity,- \(a\) is the acceleration,- \(t\) is the time.By substituting known values, we calculated a travel distance of 63.88 meters during deceleration.Furthermore, to find the time needed to stop completely, we rearranged another kinematic equation:\[v_f = v_i + at\]These equations are powerful as they allow us to solve various motion-related problems systematically.

Conversion of units

Correct conversion of units is vital in solving physics problems accurately. In our exercise, we started with speeds given in kilometers per hour (km/h), but needed them in meters per second (m/s). To convert, use the factor:\[1 \, \text{km/h} = \frac{1}{3.6} \, \text{m/s}\]Applying this, the initial speed of 90 km/h converts to 25 m/s, and the final speed of 25 km/h converts to 6.94 m/s.Understanding how to convert units ensures that all values are in coherent units before applying any formulas.It's a necessary skill for solving problems correctly, as it prevents the mixing of incompatible units.