Question

You drive with a constant speed of \(13.5 \mathrm{~m} / \mathrm{s}\) for \(30.0 \mathrm{~s}\). You then steadily accelerate for \(10.0 \mathrm{~s}\) to a speed of \(22.0 \mathrm{~m} / \mathrm{s}\). You then slow smoothly to a stop in \(10.0 \mathrm{~s} .\) How far have you traveled?

Short answer

Answer: The vehicle traveled a total of 692.5 meters.

Step-by-step solution

Stage 1 - Constant speed

During the first stage, the vehicle travels at a constant speed of \(13.5 \mathrm{~m} / \mathrm{s}\) for \(30.0 \mathrm{~s}\). To find the distance traveled in this stage, we can use the formula for distance: $$d = vt$$ where d represents distance, \(v\) represents speed, and \(t\) represents time. So, in this case, we get: $$d_1 = (13.5 \mathrm{~m} / \mathrm{s})(30.0 \mathrm{~s}) = 405 \mathrm{~m}$$

Stage 2 - Acceleration

During the second stage, the vehicle accelerates from a speed of \(13.5 \mathrm{~m} / \mathrm{s}\) to \(22.0 \mathrm{~m} / \mathrm{s}\) in \(10.0 \mathrm{~s}\). First, we need to find the acceleration rate using the formula $$a = \frac{v_f - v_0}{t}$$ where \(a\) represents acceleration, \(v_f\) represents final speed, \(v_0\) represents initial speed, and \(t\) represents time. So, in this case, we get: $$a = \frac{22.0 \mathrm{~m} / \mathrm{s} - 13.5 \mathrm{~m} / \mathrm{s}}{10.0 \mathrm{~s}} = 0.85 \mathrm{~m} / \mathrm{s}^2$$ Now we can find the distance traveled in this stage using the equation $$d = v_0t +\frac{1}{2}at^2$$ We get: $$d_2 = (13.5 \mathrm{~m} / \mathrm{s})(10.0 \mathrm{~s}) +\frac{1}{2}(0.85 \mathrm{~m} / \mathrm{s}^2)(10 \mathrm{~s})^2 = 135 \mathrm{~m} + 42.5 \mathrm{~m} = 177.5 \mathrm{~m}$$

Stage 3 - Deceleration

During the third stage, the vehicle decelerates from a speed of \(22.0 \mathrm{~m} / \mathrm{s}\) to a stop in \(10.0 \mathrm{~s}\). First, we need to find the deceleration rate using the formula $$a = \frac{v_f - v_0}{t}$$ So, in this case, we get: $$a = \frac{0 \mathrm{~m} / \mathrm{s} - 22.0 \mathrm{~m} / \mathrm{s}}{10.0 \mathrm{~s}} = -2.2 \mathrm{~m} / \mathrm{s}^2$$ Now we can find the distance traveled in this stage using the same equation as before $$d = v_0t +\frac{1}{2}at^2$$ We get: $$d_3 = (22.0 \mathrm{~m} / \mathrm{s})(10.0 \mathrm{~s}) +\frac{1}{2}(-2.2 \mathrm{~m} / \mathrm{s}^2)(10 \mathrm{~s})^2 = 220 \mathrm{~m} - 110 \mathrm{~m} = 110 \mathrm{~m}$$

Total Distance

Now we can find the total distance traveled by adding the distances in each stage: $$d_T = d_1 + d_2 + d_3 = 405 \mathrm{~m} + 177.5 \mathrm{~m} + 110 \mathrm{~m} = 692.5 \mathrm{~m}$$ So, the vehicle traveled a total of \(692.5 \mathrm{~m}\).

Key concepts

Constant Speed

Understanding constant speed is crucial for solving kinematics problems. It refers to motion where the speed (the rate of change of position) remains the same over a period of time. This means the object covers equal distances in equal intervals of time without changing its velocity.

Take the example from the exercise solution where a vehicle travels at a constant speed of \(13.5 \mathrm{~m} / \mathrm{s}\) for \(30.0 \mathrm{~s}\). To calculate the distance during this phase, the formula \(d = vt\) is used, where \(d\) is distance, \(v\) is speed, and \(t\) is time. Here, no force is acting on the vehicle to accelerate or decelerate it; hence the velocity remains unchanged, leading to a calculated distance of \(405 \mathrm{~m}\).

Acceleration

Acceleration is a measure of how quickly an object's speed is changing. In kinematics, when an object’s velocity increases over time, it is said to be accelerating. Acceleration can be calculated with the formula \(a = \frac{v_f - v_0}{t}\), where \(a\) represents the acceleration, \(v_f\) is the final speed, \(v_0\) is the initial speed, and \(t\) is the time over which the change occurs.

In our problem, the vehicle accelerated from \(13.5 \mathrm{~m} / \mathrm{s}\) to \(22.0 \mathrm{~m} / \mathrm{s}\) over \(10.0 \mathrm{~s}\), resulting in an acceleration of \(0.85 \mathrm{~m} / \mathrm{s}^2\). To find the distance covered, we utilize the equation \(d = v_0t +\frac{1}{2}at^2\), which accounts for the initial velocity, time, and acceleration factor, resulting in \(177.5 \mathrm{~m}\) traveled during the acceleration phase.

Deceleration

Deceleration is the counterpart to acceleration – it signifies a decrease in speed. The principles and formulas applied to acceleration can also be used for deceleration with a critical difference: the acceleration value will be negative to indicate the reduction in speed.

In the context of our exercise, the vehicle decelerates to a stop from \(22.0 \mathrm{~m} / \mathrm{s}\) over \(10.0 \mathrm{~s}\), translating to a deceleration (negative acceleration) of \( -2.2 \mathrm{~m} / \mathrm{s}^2\). Comparably, the distance covered can be calculated using the starting speed, time, and negative acceleration, showing the vehicle traveled \(110 \mathrm{~m}\) while slowing down to a halt.

Total Distance Traveled

The total distance traveled in a multi-stage journey is simply the sum of the distances covered in each stage. It is an accumulation of constant speed, acceleration, and deceleration intervals, which together yield the complete distance an object has moved.

In our exercise, the vehicle's total travel distance is found by adding up the distances from the constant speed (\(405 \mathrm{~m}\)), acceleration (\(177.5 \mathrm{~m}\)), and deceleration (\(110 \mathrm{~m}\)) phases. Thus, the vehicle's total distance traveled amounts to \(692.5 \mathrm{~m}\), demonstrating how combining these three concepts provides an overall picture of the vehicle's motion journey.