Question

A car moving initially at a speed of $50\mathrm{mi}/\mathrm{h}(\approx 80\mathrm{km}/\mathrm{h})$ and weighing $3000\mathrm{lb}(\approx 13,000\mathrm{N})$ is brought to a stop in a distance of $200\mathrm{ft}(\approx 61\mathrm{m})$. Find $(a)$ the braking force and $(b)$ the time required to stop. Assuming the same braking force, find $(c)$ the distance and $(d)$ the time required to stop if the car were going $25\mathrm{mi}/\mathrm{h}(\approx 40\mathrm{km}/\mathrm{h})$ initially.

Short answer

The braking force is approximately 11218.79 Newtons (but considered negative if the motion's direction is set as positive). The time required to bring the car to stop from 50 mi/h is about 2.71 seconds. If the car's initial speed were 25 mi/h instead, it would require about 7.55 meters to stop and would do so in approximately 1.35 seconds.

Step-by-step solution

Convert units to the SI system

The speed should be converted from miles per hour to meters per second, the weight from pounds to Newtons, and the distance from feet to meters. Using unit conversion: speed $v1=50mi/h=22.35m/s$, weight $W=3000lb=13344.5N$, stopping distance $d1=200ft=60.96m$. The gravitational force is $g=9.8m/s^2$. The mass of the car $m=W/g=1360.46kg$.

Find the braking force

Braking force can be obtained from the equation of motion: $v^2=u^2-2ax$, where $a$ is the deceleration, $u$ is the initial velocity, $v$ is the final velocity, and $x$ is the displacement(or distance travelled). Solve this formula for $a=(u^2-v^2)/(2\ast x)$. The result is $a=-{22.35}^2/(2\ast 60.96)=-8.25m/s^2$. The negative sign represents braking or slowing down. The force required for this deceleration is $f=m\ast a=1360.46\ast -8.25=-11218.79N$. By law of motion, the force of braking is exerted in the opposite direction to the motion, hence the negative sign.

Find the time needed to stop

Time can be obtained using the first equation of motion: $v=u+at$, where $t$ is the time, rearranging for $t$ gives $t=(v-u)/a=(0-22.35)/-8.25=2.71seconds$.

Assuming the same braking force, find the distance and time required to stop if the car were going 25 mi/h initially

Repeat Step 1, but for $u2=25mi/h=11.176m/s$. Maintaining the same deceleration $a=-8.25m/s^2$, apply the equation of motion to find the new stopping distance $x=(u^2-v^2)/(2a)=({11.176}^2-0)/(2\ast -8.25)=7.55m$. Then use the first equation of motion to find the new stopping time: $t=(v-u)/a=(0-11.176)/-8.25=1.35seconds$.

Key concepts

Equations of Motion

Equations of motion form the backbone of solving various physics problems related to objects in motion. These mathematical formulas help us understand how an object moves under the influence of forces.
The three key equations of motion in physics are crucial when determining variables like velocity, acceleration, distance, and time. Here they are:

• First equation: $v=u+at$ - This equation helps find the final velocity $v$, given the initial velocity $u$, acceleration $a$, and time $t$.
• Second equation: $s=ut+\frac{1}{2}a{t}^2$ - This equation is used to calculate the distance $s$ traveled by an object.
• Third equation: $v^2=u^2+2as$ - It facilitates finding either the velocity or distance if the other elements are known.

By mastering these equations, you can unravel many dynamics problems, like calculating stopping distances or times for a braking vehicle.

Deceleration

Deceleration, often referred to as negative acceleration, is the rate at which an object slows down. When an object like a car brakes, it experiences deceleration.
Mathematically, deceleration follows similar formulas to acceleration but results in a negative value to indicate a reduction in speed. For vehicles, this deceleration is typically due to the frictional force applied by brakes, which counteracts the forward motion.
To calculate deceleration using equations of motion:

• Substitute the final velocity $v$ as zero (for a complete stop) and rearrange the third equation of motion: $a=\frac{u^2-v^2}{2x}$.
• The negative value obtained indicates slowing down - the essence of deceleration.

Understanding deceleration is vital not just in physics problems, but in real-life applications like ensuring vehicles brake within safe distances.

Unit Conversion

Unit conversion is a frequent requirement when dealing with physics problems, especially those involving motion. This process ensures all quantities are in compatible units, typically the International System of Units (SI), which simplifies calculations.
For example, converting miles per hour to meters per second aids in calculating speeds in equations of motion. Similarly, converting weight from pounds to Newtons (using the gravitational force) helps in determining forces accurately.

• Speed conversion: $1\text{ mi/h}=0.44704\text{ m/s}$.
• Weight conversion: Using $1\text{ lb}=4.44822\text{ N}$.
• Distance conversion: $1\text{ ft}=0.3048\text{ m}$.

Performing these conversions at the start of solving a physics problem guarantees that all units align correctly—leading to accurate outcomes.