Motion at Constant Acceleration - Examples

Question: You are designing an airport for small planes. One kind of airplane that might use this airfield must reach a speed before takeoff of at least v and can accelerate at a constant rate of a.

(a) If the runway is 150 m long, can this airplane reach the required speed for takeoff?

(b) If not, what minimum length must the runway have?

Note: The numerical values of the required speed and acceleration are missing from the supplied question, so the solution below is expressed in terms of v and a.

Since the airplane starts from rest and accelerates uniformly, use the equation that connects velocity, acceleration, and displacement without requiring time:

Since u = 0:

The minimum runway length needed to reach takeoff speed v is:

Compare this required length with the available 150 m.

Case 1: If v²/(2a) ≤ 150, the airplane can reach the required takeoff speed within the runway.

Case 2: If v²/(2a) > 150, the runway is too short. The minimum required runway length is v²/(2a).

Find the maximum speed the airplane can reach over 150 m. Use:

Compare this speed with the required takeoff speed v.

The airplane must accelerate from rest to the required takeoff speed v. From:

Rearranging:

Substitute the numerical takeoff speed and acceleration from the complete textbook question to obtain the runway length in metres.

For constant acceleration, the average velocity is:

Since the airplane starts from rest, u = 0:

Distance travelled equals average velocity multiplied by time:

Using v = at, the time to reach the takeoff speed is:

Substitute into the distance equation:

This gives the same minimum runway length as Method 1.

First calculate how long the airplane needs to reach takeoff speed:

Since u = 0:

Then use the displacement equation:

Since u = 0:

Question: How long does it take a car to cross a 30.0-m-wide intersection after the light turns green, if the car accelerates from rest at a constant 2.00 m/s²?

For constant acceleration, the displacement equation is:

Substitute the given values:

First, calculate the final velocity using:

Since u = 0:

For constant acceleration, average velocity is:

Because u = 0 and v = at:

Displacement equals average velocity multiplied by time:

The third equation is:

Substitute u = 0, a = 2.00 m/s², and s = 30.0 m:

Now use the first equation of motion:

The car starts from rest and accelerates uniformly. Its final speed after crossing the intersection is approximately 10.95 m/s.

Its average velocity is:

Therefore:

Question: A car starts from rest and accelerates at a constant 10 m/s² during a ¼-mile (402 m) race. How fast is the car going at the finish line?

Given Data

• Initial velocity: u = 0 m/s (starts from rest)
• Constant acceleration: a = 10 m/s²
• Displacement: s = 402 m
• Final velocity: v = ?

Method 1: Equation of Motion

Since the initial velocity, acceleration and displacement are known, use the third equation of motion:

Substitute the given values:

Rounding to the nearest whole number:

Method 2: Using Average Velocity

For constant acceleration, the average velocity is the arithmetic mean of the initial and final velocities:

Using displacement = average velocity × time:

s = [(u + v)/2]t

Because u = 0:

s = vt/2, so v = 2s/t.

From the first equation of motion, v = at, hence t = v/a.

Substitute this into the displacement equation:

s = v(v/a)/2 = v²/(2a)

Therefore:

Method 3: Using a Velocity–Time Graph

Since acceleration is constant and the car starts from rest, its velocity increases uniformly from 0 to v. The area under the velocity–time graph gives the displacement.

The graph is a triangle, so:

Using v = at, we get t = v/10.

Substitute:

Method 4: Using the Distance–Time Equation

For motion from rest with constant acceleration:

Since u = 0:

402 = ½ × 10 × t²

402 = 5t²

t² = 80.4

t = √80.4 ≈ 8.97 s

Now use v = u + at:

v = 0 + 10 × 8.97

Final Answer

EXAMPLE 2–9   ESTIMATE   Braking distances. Estimate the minimum stopping distance for a car, which is important for traffic safety and traffic design. The problem is best dealt with in two parts, two separate time intervals. (1) The first time interval begins when the driver decides to hit the brakes, and ends when the foot touches the brake pedal. This is the “reaction time” during which the speed is constant, so a = 0. (2) The second time interval is the actual braking period when the vehicle slows down (a ≠ 0) and comes to a stop. The stopping distance depends on the reaction time of the driver, the initial speed of the car (the final speed is zero), and the deceleration of the car. For a dry road and good tires, good brakes can decelerate a car at a rate of about 5 m/s² to 8 m/s². Calculate the total stopping distance for an initial velocity of 50 km/h (= 14 m/s ≈ 31 mi/h) and assume the acceleration of the car is −6.0 m/s² (the minus sign appears because the velocity is taken to be in the positive x direction and its magnitude is decreasing). Reaction time for normal drivers varies from perhaps 0.3 s to about 1.0 s; take it to be 0.50 s.

Initial velocity: u = 14 m/s

Final velocity: v = 0 m/s

Braking acceleration: a = −6.0 m/s²

Reaction time: tᵣ = 0.50 s

Required: Total stopping distance

The complete stopping process consists of two separate time intervals.

Part A: Reaction Distance

During the reaction time, the speed remains constant. Therefore, distance is equal to velocity multiplied by time.

Substitute the values:

Part B: Braking Distance

During braking, the initial velocity is 14 m/s, the final velocity is zero, and acceleration is −6.0 m/s².

Use the third equation of motion:

Substitute the values:

Part C: Total Stopping Distance

First calculate the time taken by the car to stop after the brakes are applied.

Use the first equation of motion:

Substitute the values:

For constant acceleration, average velocity during braking is:

Braking distance is average velocity multiplied by braking time:

Add the reaction distance:

We can also calculate the braking distance directly from the second equation of motion:

We already know that the braking time is: tᵦ = 14/6 s.

Substitute the values:

Therefore:

The braking distance for constant deceleration can be calculated directly using the magnitude of acceleration.

The total stopping distance is the sum of the reaction distance and the braking distance.

Substitute the values:

The reaction distance is about 7 m because the car travels at 14 m/s for half a second.

During braking, the speed decreases uniformly from 14 m/s to zero. Its average speed is 7 m/s, and the braking time is approximately 2.33 s.

Therefore, the braking distance is approximately:

Adding both parts gives a total stopping distance of about 23 m.

If the initial speed doubles, while reaction time and braking acceleration remain unchanged, reaction distance doubles and braking distance becomes four times as large.