Motion at Constant Acceleration - Problems - 2

Given Data

• Initial velocity: u = 18.0 m/s
• Final velocity for stopping: v = 0 m/s
• Deceleration magnitude: 3.65 m/s²
• Reaction time: tᵣ = 0.350 s
• Distance to intersection: 20.0 m

Understanding the Motion

The car travels in two stages:

• Reaction stage: The driver has seen the red light but has not yet applied the brakes. The car continues at 18.0 m/s.
• Braking stage: The brakes are applied, and the car decelerates at 3.65 m/s² until it stops.

Step 1: Calculate the Reaction Distance

During the reaction time, the velocity remains constant. Therefore, use:

dᵣ = 18.0 × 0.350

dᵣ = 6.30 m

The car travels 6.30 m before the brakes are applied. The remaining distance to the intersection is:

d remaining = 20.0 − 6.30

Step 2: Calculate the Braking Distance

Use the third equation of motion because the initial velocity, final velocity, and acceleration are known.

Take the direction of the car’s motion as positive. During braking, a = −3.65 m/s².

0² = (18.0)² + 2(−3.65)dᵦ

0 = 324 − 7.30dᵦ

7.30dᵦ = 324

dᵦ = 324/7.30

Step 3: Calculate the Total Stopping Distance

d total = dᵣ + dᵦ

d total = 6.30 + 44.38

Step 4: Will the Car Stop Before the Intersection?

The car is only 20.0 m from the intersection, but it needs approximately 50.68 m to stop.

50.68 m > 20.0 m

Step 5: Find How Far the Car Travels Beyond the Intersection

Choose the position where the driver first sees the red light as x = 0. The beginning of the intersection is then at x = 20.0 m.

The car stops after travelling 50.68 m from its initial position. Therefore, its final position relative to the beginning of the intersection is:

x final relative to intersection = 50.68 − 20.0

The positive result means the car stops beyond the beginning of the intersection, in the original direction of travel.

Alternative Method: Calculate the Velocity at the Intersection

The car has 13.70 m remaining when the brakes are applied. Use the third equation of motion to find its speed after braking over this distance:

v² = (18.0)² + 2(−3.65)(13.70)

v² = 324 − 100.01

v = √223.99

v ≈ 14.97 m/s

Since the velocity is still positive at the intersection, the car is still moving forward when it reaches it. It therefore cannot stop in time.

Question

(II) A 75-m-long train begins uniform acceleration from rest. The front of the train has a speed of 18 m/s when it passes a railway worker who is standing 180 m from where the front of the train started. What will be the speed of the last car as it passes the worker? (See Fig. 2–38.)

Step 1: Find the acceleration of the train

Since the train starts from rest and accelerates uniformly, use the third equation of motion:

Substitute the known values:

Step 2: Find the distance travelled by the front when the last car reaches the worker

The worker is 180 m from the train’s starting point. For the last car to pass the worker, the front of the train must travel an additional distance equal to the train’s length, 75 m.

Step 3: Find the speed of the train at 255 m

Apply the third equation of motion again, using the same acceleration and initial velocity:

Question

(II) A space vehicle accelerates uniformly from 85 m/s at t = 0 to 162 m/s at t = 10.0 s. How far did it move between t = 2.0 s and t = 6.0 s?

Method 1: Use the equations of motion

Step 1: Calculate the constant acceleration using the first equation of motion.

Step 2: Calculate the displacement from t = 0 to t = 2.0 s.

Using the second equation of motion:

Step 3: Calculate the displacement from t = 0 to t = 6.0 s.

Step 4: Find the displacement between t = 2.0 s and t = 6.0 s by subtracting the two positions.

Method 2: Use average velocity over the time interval

For uniform acceleration, average velocity is the arithmetic mean of the initial and final velocities for the interval.

Calculate the average velocity:

The elapsed time is:

Question

(II) A runner hopes to complete the 10,000-m run in less than 30.0 min. After running at constant speed for exactly 27.0 min there are still 1200 m to go. The runner must then accelerate at 0.20 m/s² for how many seconds in order to achieve the desired time?

Step 1: Calculate the remaining time

The runner wants to finish the race in 30.0 min and has already run for 27.0 min.

Step 2: Find the runner’s initial speed

Before accelerating, the runner covers 10,000 − 1200 = 8800 m in 1620 s at constant speed.

Using the definition of constant speed:

Step 3: Use the equation of motion

Let t be the number of seconds the runner accelerates at 0.20 m/s². The runner covers 1200 m during the remaining 180 s, so the motion during this interval satisfies:

Step 4: Solve the quadratic equation

Using the quadratic formula, t = [−b ± √(b² − 4ac)] / 2a:

The negative root represents a negative time and is not physically meaningful, so we reject it.

Step 5: Check whether the target time is achieved

The runner accelerates for approximately 85.7 s and then runs at the resulting speed for the rest of the 180 s interval.

The remaining time at the new speed is:

Question

(III) A fugitive tries to hop on a freight train traveling at a constant speed of 5.0 m/s. Just as an empty box car passes him, the fugitive starts from rest and accelerates at a = 1.4 m/s² to his maximum speed of 6.0 m/s, which he then maintains. (a) How long does it take him to catch up to the empty box car? (b) What is the distance traveled to reach the box car?

Step 1: Calculate the time needed to reach maximum speed

The fugitive accelerates from rest until his speed reaches 6.0 m/s. Using the first equation of motion:

Step 2: Find the distance travelled by the fugitive during acceleration

Use the second equation of motion:

Step 3: Find the distance travelled by the train during this time

The train continues moving at a constant speed of 5.0 m/s.

At the end of the acceleration phase, the train is ahead of the fugitive by:

Step 4: Calculate the time needed to close the remaining gap

After reaching 6.0 m/s, the fugitive maintains that speed. The train continues at 5.0 m/s. Therefore, the fugitive gains on the train at the relative speed:

The time to close the remaining gap is:

The total time from the start until the fugitive catches the box car is:

Step 5: Calculate the total distance travelled by the fugitive

The fugitive travels 12.86 m while accelerating and then travels at 6.0 m/s for another 8.57 s.

Hence, the total distance travelled is:

Question (III)

Mary and Sally are in a foot race (Fig. 2–39). When Mary is 22 m from the finish line, she has a speed of 4.0 m/s and is 5.0 m behind Sally, who has a speed of 5.0 m/s. Sally thinks she has an easy win and so, during the remaining portion of the race, decelerates at a constant rate of 0.40 m/s² to the finish line. What constant acceleration does Mary now need during the remaining portion of the race, if she wishes to cross the finish line side-by-side with Sally?

Step 1: Find Sally’s remaining distance

Mary is 22 m from the finish line, and Sally is 5.0 m ahead. Therefore, Sally has only

Step 2: Calculate the time Sally takes to reach the finish

Since Sally decelerates uniformly, use the equation of motion:

Substitute Sally’s values:

Using the quadratic formula:

Sally reaches the finish line at the first time, so the physically relevant time is:

Step 3: Calculate Mary’s acceleration

To finish side-by-side, Mary must cover 22 m in the same time, 4.06 s. Her initial speed is 4.0 m/s.

Apply the equation of motion:

Using the unrounded value of the time gives a_M ≈ 0.698 m/s².

Question (III)

An unmarked police car traveling a constant 95 km/h is passed by a speeder traveling 135 km/h. Precisely 1.00 s after the speeder passes, the police officer steps on the accelerator; if the police car’s acceleration is 2.60 m/s², how much time passes before the police car overtakes the speeder (assumed moving at constant speed)?

Given data

Step 1: Convert speeds into m/s

Use the conversion formula:

For the police car:

For the speeder:

Step 2: Find the initial lead of the speeder

During the first 1.00 s, the police car continues at its constant speed while the speeder moves faster. Therefore, the speeder gains a distance equal to the difference between their speeds multiplied by the reaction time.

Step 3: Write the equations of motion

Take the position where the speeder passes the police car as x = 0 at t = 0. Let t be the time measured from that instant.

For the speeder, which moves at constant speed:

For the police car, the motion has two stages. During the first second, it moves at constant speed. After that, it accelerates.

For t ≥ 1.00 s, its position is

Step 4: Find when the police car catches the speeder

At the overtaking instant, both vehicles have the same position:

Applying the quadratic formula gives two mathematical roots. The smaller root is less than 1.00 s, before the police car begins accelerating, so it is not physically valid for this stage. The valid root is approximately

Step 5: Interpret the time

The value t = 10.45 s is measured from when the speeder first passes the police car. The officer waits 1.00 s before accelerating.