Problem 15 - Basic Engineering Mathematics

Question

The distance \(s\) metres from a fixed point of a vehicle travelling in a straight line with constant acceleration, \(a \mathrm{~m} / \mathrm{s}^{2}\), is given by \(s=u t+\frac{1}{2} a t^{2}\), where \(u\) is the initial velocity in \(\mathrm{m} / \mathrm{s}\) and \(t\) the time in seconds. Determine the initial velocity and the acceleration given that \(s=42 \mathrm{~m}\) when \(t=2 \mathrm{~s}\) and \(s=144 \mathrm{~m}\) when \(t=4 \mathrm{~s}\). Find also the distance travelled after \(3 \mathrm{~s}\).

Step-by-Step Solution

Answer

Initial velocity: 6 m/s, acceleration: 15 m/s², distance after 3s: 85.5 m.

Step 1: Write the two equations

We begin by substituting the given values into the formula \( s = ut + \frac{1}{2} at^2 \) for the two different sets of conditions. For \( t = 2 \), we get the equation \( 42 = 2u + 2a \). For \( t = 4 \), we get \( 144 = 4u + 8a \). These are our two equations: \( 42 = 2u + 2a \) and \( 144 = 4u + 8a \).

Step 2: Solve the system of equations

We will solve the two equations simultaneously. Simplify the first equation: \( 42 = 2(u + a) \), so \( u + a = 21 \). Simplify the second: \( 144 = 4(u + 2a) \), so \( 4u + 8a = 144 \) as is. Divide the second equation by 4: \( u + 2a = 36 \). Now solve \( u + a = 21 \) and \( u + 2a = 36 \). Subtract to eliminate \( u: (u + 2a) - (u + a) = 36 - 21 \). This gives \( a = 15 \).

Step 3: Substitute to find initial velocity

With \( a = 15 \), substitute back into \( u + a = 21 \). Thus, \( u + 15 = 21 \) resulting in \( u = 6 \). So, the initial velocity is \( 6 \text{ m/s} \).

Step 4: Calculate the distance traveled in 3 seconds

Using the formula \( s = ut + \frac{1}{2} at^2 \) with \( u = 6 \), \( a = 15 \), and \( t = 3 \), calculate: \( s = 6(3) + \frac{1}{2}(15)(3^2) \). This is \( s = 18 + \frac{1}{2}(15)(9) \), which simplifies to \( s = 18 + 67.5 \), resulting in \( s = 85.5 \text{ m} \).

Key Concepts

Constant Acceleration

In kinematics, constant acceleration means that the rate of change of velocity with respect to time stays the same throughout the motion. When an object moves with constant acceleration, it simplifies the analysis of motion because you can use straightforward mathematical formulas. For instance, the distance traveled in such a scenario can be determined using the equation \[ s = ut + \frac{1}{2} at^2 \]. Here:

• \( s \) represents the distance covered,
• \( u \) is the initial velocity of the moving object,
• \( a \) is the constant acceleration,
• \( t \) symbolizes the time period over which the motion occurs.

This formula is derived from integrating the velocity function over time. The constant acceleration ensures that the phenomenon can be modeled accurately and predictions can easily be made.

Initial Velocity

The initial velocity \( u \) is a crucial component in analyzing an object’s motion. Initial velocity refers to the speed and direction at which an object starts its journey along a path. In the equation \( s = ut + \frac{1}{2} at^2 \), it affects the overall distance traveled.
To find the initial velocity in an equation involving multiple known variables, such as distance or time, one can isolate \( u \) when information about the acceleration and the trajectory of the object is provided.
In our exercise, determining the initial velocity required setting up a system of equations, allowing us to solve for \( u \) by first finding the value of the constant acceleration \( a \). Once \( a \) was identified as 15 m/s², we concluded that the initial velocity \( u \) was 6 m/s.

Distance Travelled

Understanding the distance traveled in a particular timeframe is fundamental to solving kinematic problems. The distance or displacement \( s \) depends on initial velocity, time, and acceleration in constant acceleration scenarios.
Employing the kinematic equation \[ s = ut + \frac{1}{2} at^2 \], allows for the calculation of how far an object travels under these conditions.
In the solution, after determining both the initial velocity and acceleration, we were able to predict the distance covered by the object at any given time. For instance, after 3 seconds, the distance reached was calculated to be 85.5 meters.

System of Equations

A system of equations is an essential mathematical tool in solving problems where multiple unknowns are involved. In this context, a system of equations was used to determine both the initial velocity and the acceleration.
Initially, two different sets of conditions from the motion were provided, which allowed us to form two equations:

• \( 42 = 2u + 2a \)
• \( 144 = 4u + 8a \)

Solving these equations simultaneously helps in eliminating one variable, simplifying the process to find the other.
By simplifying and subtracting, we found the acceleration, \( a = 15 \text{ m/s}^2 \). Subsequently, this information was substituted back to uncover the initial velocity \( u = 6 \text{ m/s} \). Using these methods, you can effectively handle various kinematics problems.