Question

How much time does it take for a car to accelerate from a standing start to \(22.2 \mathrm{~m} / \mathrm{s}\) if the acceleration is constant and the car covers \(243 \mathrm{~m}\) during the acceleration?

Short answer

Answer: The car takes approximately 21.89 seconds to accelerate from a standing start to 22.2 m/s while covering 243 meters.

Step-by-step solution

List given values

We have the following information: - Initial velocity, \(v_1 = 0 \mathrm{~m} / \mathrm{s}\) (since the car starts from a standing start, initial velocity is 0) - Final velocity, \(v_2 = 22.2 \mathrm{~m} / \mathrm{s}\) - Distance covered, \(d = 243 \mathrm{~m}\) We need to find the time taken for this acceleration, \(t\).

Determine acceleration using the 1st equation of motion

We know the equation \(v_2 = v_1 + at\), where \(a\) is the constant acceleration and \(t\) is the time taken. Since \(v_1 = 0\), the equation becomes \(v_2 = at\). We can rewrite this equation to find the acceleration as a function of time and final velocity: $$ a = \frac{v_2}{t} $$

Determine time using the 2nd equation of motion

Now we can use the second equation of motion, \(d = v_1t + \frac{1}{2}at^2\). Since \(v_1 = 0\), the equation becomes: $$ d = \frac{1}{2}at^2 $$

Substitute the acceleration value into the 2nd equation of motion

From step 2, we have \(a = \frac{v_2}{t}\). Substituting this value into the equation from step 3, we get: $$ d = \frac{1}{2}\left(\frac{v_2}{t}\right)t^2 $$

Solve for time, t

Now we have an equation with only one unknown, \(t\). We can solve for time by simplifying and isolating \(t\): $$ d = \frac{1}{2}\frac{v_2t^2}{t} $$ Multiplying both sides by \(2t\), we get: $$ 2td = v_2t^2 $$ Divide both sides by \(v_2\) and \(t\): $$ t = \frac{2d}{v_2} $$

Calculate Time

Now, we can plug in the given values for distance and final velocity into the equation to find the time: $$ t = \frac{2(243\mathrm{~m})}{22.2\mathrm{~m}/\mathrm{s}} $$ $$ t = \frac{486\mathrm{~m}}{22.2\mathrm{~m}/\mathrm{s}} $$ $$ t = 21.89 \mathrm{~s} $$ The car takes approximately 21.89 seconds to accelerate from a standing start to 22.2 m/s while covering 243 meters.

Key concepts

Equations of Motion

When studying the motion of objects, we use equations of motion to predict and analyze the trajectory and velocity of objects moving under the influence of constant acceleration. These fundamental formulas link the movement of an object with time, velocity, acceleration, and displacement. The first equation of motion, often written as \(v = u + at\), associates the final velocity \(v\) with the initial velocity \(u\), acceleration \(a\), and time \(t\). This relationship provides us with the foundation for understanding how an object’s velocity changes over time when subjected to uniform acceleration.

Another crucial equation is \(s = ut + \frac{1}{2}at^2\), where \(s\) stands for the displacement. By using these equations, we can solve various kinematic problems, like determining how much time a car takes to accelerate to a certain speed or the distance it covers in that time. It connects acceleration to displacement, and in the context of our car problem, it gave us a way to compute the time taken to reach the final velocity while covering a specific distance.

Constant Acceleration

In kinematics problems, the assumption of constant acceleration is crucial because it simplifies the equations used. Constant acceleration means the rate of change of velocity of an object is uniform over time. This is often a reasonable approximation for many real-life situations, such as a car accelerating on a straight road. It implies that the object in question doesn't speed up or slow down at different rates during the period of observation, but rather changes its velocity by the same amount every second.

The concept of constant acceleration is deeply embedded in the equations of motion. It allows us to predict the future state of a moving object based on its current state without having to consider complex factors that might cause variable acceleration. In the textbook exercise, assuming constant acceleration lets us accurately calculate how long the car takes to reach a certain speed.

Solving for Time

In kinematic equations, solving for time involves isolating and calculating the variable \(t\) to understand how long an object will take to change from one state of motion to another under constant acceleration. This aspect of motion can be crucial for planning and predicting the outcomes of physical events. The ability to solve for time is not only applicable in physics but also in daily practical situations like estimating travel times or athletic performances.

Using the correct rearrangement of the equations of motion, as demonstrated in the exercise, we can isolate the variable \(t\) to calculate the time it takes for the object (in this case, a car) to move from its initial velocity to its final velocity. The process usually involves algebraic manipulation to get the time on one side of the equation and then plugging in the known values to find the specific duration.

Initial and Final Velocity

Initial velocity refers to the velocity of an object before it experiences acceleration. In the case of the car from the textbook exercise, the initial velocity was zero since it started from a standing start. Final velocity, on the other hand, is the velocity at which the object is moving after it has accelerated. It represents the new state of motion after the influence of a force.

Understanding both initial and final velocities is key to solving kinematics problems because these values help to define the 'before' and 'after' snapshots of motion. In our exercise, the final velocity was the target value that we wanted the car to reach. Using the respective kinematic equations, we can see how initial velocity, final velocity, acceleration, and time are intertwined, allowing us to describe motion fully. Whether it's for planning purposes, like determining speed limits, or for understanding natural phenomena, knowing these velocities is essential.