Question

At launch a rocket ship weighs 4.5 million pounds. When it is launched from rest, it takes 8.00 s to reach 161 km/h; at the end of the first 1.00 min, its speed is 1610 km/h. (a) What is the average acceleration (in m/s${}^2$) of the rocket (i) during the first 8.00 s and (ii) between 8.00 s and the end of the first 1.00 min? (b) Assuming the acceleration is constant during each time interval (but not necessarily the same in both intervals), what distance does the rocket travel (i) during the first 8.00 s and (ii) during the interval from 8.00 s to 1.00 min?

Short answer

(a) (i) 5.59 m/s², (ii) 7.75 m/s²; (b) (i) 179.2 m, (ii) 12226.4 m.

Step-by-step solution

Convert Units for Velocity

Convert the initial and final velocities from km/h to m/s. The conversion factor is 1 km/h = $\frac{1}{3.6}$ m/s.**Initial Velocity during first 8 s:**$$v_1=161\text{km/h}=\frac{161}{3.6}\approx 44.72\text{m/s}$$**Final Velocity at 1 min:$$v_2=1610\text{km/h}=\frac{1610}{3.6}\approx 447.22\text{m/s}$$

Calculate Average Acceleration for First 8 Seconds

The average acceleration is given by the formula:$$a=\frac{\mathrm{\Delta }v}{\mathrm{\Delta }t}$$For the first 8 seconds, initial velocity is 0, and final velocity is 44.72 m/s. Time is 8 s.$$a_1=\frac{44.72\text{m/s}-0\text{m/s}}{8\text{s}}=5.59{\text{m/s}}^2$$

Calculate Average Acceleration between 8s and 1 min

The time for this interval is 52 seconds (60 - 8).Use the velocities calculated previously:Initial velocity at 8 s is 44.72 m/s and final velocity at 1 min is 447.22 m/s.$$a_2=\frac{447.22\text{m/s}-44.72\text{m/s}}{52\text{s}}=7.75{\text{m/s}}^2$$

Calculate Distance Traveled During First 8 Seconds

Use the equation for distance under constant acceleration:$$d=v_{i}t+\frac{1}{2}a{t}^2$$For the first 8 seconds:- Initial velocity, $v_i=0$- Acceleration, $a_1=5.59{\text{m/s}}^2$- Time, $t=8\text{s}$$$d_1=0+\frac{1}{2}\times 5.59{\text{m/s}}^2\times (8\text{s}{)}^2=179.2\text{m}$$

Calculate Distance Traveled from 8s to 1 min

Again, use the formula for distance:$$d=v_{i}t+\frac{1}{2}a{t}^2$$For this interval:- Initial velocity at 8s, $v_i=44.72\text{m/s}$- Acceleration, $a_2=7.75{\text{m/s}}^2$- Time during this interval is 52 seconds.$$d_2=44.72\text{m/s}\times 52\text{s}+\frac{1}{2}\times 7.75{\text{m/s}}^2\times (52\text{s}{)}^2=12226.4\text{m}$$

Key concepts

Average Acceleration

When discussing rocket motion, understanding acceleration helps us predict how quickly things change speed. **Average acceleration** is one of these important measures. It tells us how much the velocity of an object changes over a certain time period.

To calculate average acceleration, use the formula:

• $$a_{avg}=\frac{\mathrm{\Delta }v}{\mathrm{\Delta }t}$$

Where $\mathrm{\Delta }v$ is the change in velocity and $\mathrm{\Delta }t$ is the time it takes for that change.

In the context of our rocket exercise, we calculated average acceleration in two time segments:

• The first 8 seconds, where acceleration was $5.59{\text{m/s}}^2$.
• From 8 seconds to 1 minute, where it was $7.75{\text{m/s}}^2$.

Both of these are vital for understanding the rocket's journey into space.

Constant Acceleration

Constant acceleration is a concept frequently used in physics to simplify calculations. It means the acceleration is steady and doesn’t change over time. In the rocket problem, assuming constant acceleration allows us to use specific formulas to find out how far the rocket has traveled.

When acceleration is constant, the equations of motion can be applied, like

• $$d=v_{i}t+\frac{1}{2}a{t}^2$$

where $d$ is distance, $v_i$ is the initial velocity, $a$ is acceleration, and $t$ is time.

For our rocket, seeing acceleration as constant gave us:

• 179.2 meters of travel during the first 8 seconds.
• 12,226.4 meters from 8 seconds to 1 minute.

This simplification is extremely helpful when dealing with complex real-world scenarios.

Distance Traveled

Calculating the **distance traveled** is another crucial task when analyzing rocket performance. This involves determining how much ground the rocket covers during each stage of its ascent.

The formula used to calculate distance under constant acceleration is:

• $$d=v_{i}t+\frac{1}{2}a{t}^2$$

This formula accounts for both initial velocity and the additional distance covered due to acceleration.

By applying this to our exercise:

• With the rocket starting from rest, it covered 179.2 meters in the first 8 seconds.
• Between 8 seconds and 1 minute, it traveled an impressive 12,226.4 meters.

This information helps us map out the rocket's path and understand its motion dynamics.

Velocity Conversion

Understanding the motion of a rocket often starts with converting its speed into more convenient units for calculations. **Velocity conversion** is vital because different units suit different aspects of physics.

In this problem, velocities were given in kilometers per hour (km/h). To perform calculations in meters per second (m/s) - the preferred unit in physics - we use:

• 1 km/h = $\frac{1}{3.6}$ m/s

This is because there are 3,600 seconds in an hour and 1,000 meters in a kilometer.

For example:

• Converting the rocket's initial speed of 161 km/h to m/s gave 44.72 m/s.
• Its speed at 1 minute of 1610 km/h converted to 447.22 m/s.

Converting velocities ensures all calculations are precise and based on standard units, crucial for accurate physics analyses.