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A truck starts from rest and moves with a constant acceleration of \(5.0 \mathrm{~m} / \mathrm{s}^{2}\). Find its speed and the distance traveled after \(4.0 \mathrm{~s}\) has elapsed.

Short Answer

Expert verified
The truck reaches a speed of 20 m/s and travels 40 meters in 4.0 seconds.

Step by step solution

01

Identify the Given Values

The problem provides several key values: initial speed (\(u\)) of the truck is \(0\) m/s (since it starts from rest), acceleration (\(a\)) is \(5.0\, \mathrm{m/s}^2\), and the time (\(t\)) is \(4.0\, \mathrm{s}\).
02

Use the Formula to Find the Final Speed

To find the final speed, use the formula \(v = u + at\). Substitute the given values: \(v = 0 + 5.0 \times 4.0 = 20 \mathrm{~m/s}\). So, the final speed after \(4.0\, \mathrm{s}\) is \(20\, \mathrm{m/s}\).
03

Use the Formula to Find the Distance Traveled

To calculate the distance traveled, use the formula \(s = ut + \frac{1}{2} a t^2\). Substituting the given values, we get \(s = 0 \times 4.0 + \frac{1}{2} \times 5.0 \times (4.0)^2\). This simplifies to \(s = 0 + 0.5 \times 5.0 \times 16.0\), resulting in \(s = 40 \mathrm{~m}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Constant Acceleration
In physics, constant acceleration refers to a situation where an object's acceleration does not change with time. This means that the rate at which the velocity of the object increases remains steady. If a truck, for example, starts from rest—meaning its initial speed is zero—and begins to move with a constant acceleration, its speed increases steadily each second. This constant acceleration results in a linear increase in velocity over time.

Here are some key points about constant acceleration:
  • The velocity of the object increases by the same amount every second.
  • The object's speed and distance traveled are predictable over time.
  • Constant acceleration is common in physics problems, making it an essential concept to understand.
Understanding constant acceleration helps in solving problems involving moving objects, providing a clear framework for calculating speed and distance traveled.
Kinematics Equations
Kinematics equations are a set of four equations in physics that describe the motion of objects under constant acceleration. They allow us to calculate different aspects of motion, such as speed, distance, and time.

The main kinematics equations you'll encounter are:
  • \( v = u + at \)
  • \( s = ut + \frac{1}{2} a t^2 \)
  • \( v^2 = u^2 + 2as \)
  • \( s = vt - \frac{1}{2} a t^2 \)
In these equations:
  • \( v \) is the final velocity.
  • \( u \) is the initial velocity.
  • \( a \) is the acceleration.
  • \( t \) is the time.
  • \( s \) is the distance traveled.
The kinematics equations are derived from basic principles of motion and help in solving various physics problems. When you're given three known values (such as initial speed, acceleration, and time), you can use these equations to find the unknown variables.
Distance and Speed Calculation
To solve problems involving speed and distance, especially in scenarios of constant acceleration, we use specific formulas. In the given exercise, we needed to find both the speed and the distance the truck traveled.

### Calculating Speed
We used the formula \( v = u + at \). This tells us how fast the truck is moving after a certain time if it starts from a rest (zero initial speed).

The calculation involves:
  • Identifying the initial speed \( u = 0 \) m/s, since the truck starts from rest.
  • Using the given acceleration \( a = 5.0 \, \mathrm{m/s}^2 \).
  • Multiplying the acceleration by the time \( t = 4.0 \, \mathrm{s} \).
The final speed comes out as \( v = 20 \, \mathrm{m/s} \).

### Calculating Distance
To find out how far the truck traveled, we use the formula \( s = ut + \frac{1}{2} a t^2 \). Since the initial speed is zero, it simplifies to \( s= \frac{1}{2} a t^2 \).

By substituting the known values:
  • Initial speed \( u = 0 \); hence, initial distance is zero.
  • The constant acceleration \( a = 5.0 \, \mathrm{m/s}^2 \).
  • Squaring the time \( (4.0)^2 \).
We calculate the total distance covered as \( s = 40 \, \mathrm{m} \).

This simple exercise shows how various kinematics equations are used to solve everyday problems, elucidating distances covered and speeds achieved under constant acceleration.

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Most popular questions from this chapter

A stone is thrown straight upward and it rises to a maximum height of \(20 \mathrm{~m}\). With what speed was it thrown? Take up as the positive \(y\) -direction. The stone's velocity is zero at the top of its path. Then \(v_{f y}=0, y=20 \mathrm{~m}, a=-9.81 \mathrm{~m} / \mathrm{s}^{2}\). (The minus sign arises because the acceleration due to gravity is always downward and we have taken up to be positive.) Use \(v_{s}^{2}=v_{i}^{2}+2 a y\) to find $$ v_{i y}=\sqrt{-2\left(-9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(20 \mathrm{~m})}=20 \mathrm{~m} / \mathrm{s} $$ Alternative Method You can check your result using the fact that the peak altitude is given by Eq. (2.9); that is, \(y_{p}=-v_{i}^{2} / 2 s\), and so \(v_{i}^{2}=-2 g_{y}\), or \(v_{i}^{2}=-2(-9,81 \mathrm{~m} / 3)(20\) and \(u_{i}=19.8 \mathrm{~m} / \mathrm{s}\), or to two significant figures, \(u_{i}=20 \mathrm{~m} / \mathrm{s}\).

An object starts from rest with a constant acceleration of \(8.00 \mathrm{~m} / \mathrm{s}^{2}\) along a straight line. Find \((a)\) the speed at the end of \(5.00 \mathrm{~s},(b)\) the average speed for the 5-s interval, and ( \(c\) ) the distance traveled in the \(5.00 \mathrm{~s}\). We are interested in the motion for the first \(5.00 \mathrm{~s}\). Take the direction of motion to be the \(+x\) -direction (that is, \(s=x\) ). We know that \(v_{i}=0, t=5.00 \mathrm{~s}\), and \(a=8.00 \mathrm{~m} / \mathrm{s}^{2}\). Because the motion is uniformly accelerated, the five motion equations apply. (a) \(v_{f x}=v_{i x}+a t=0+\left(8.00 \mathrm{~m} / \mathrm{s}^{2}\right)(5.00 \mathrm{~s})=40.0 \mathrm{~m} / \mathrm{s}\) (b) \(\quad v_{a v}=\frac{v_{i x}+v_{f x}}{2}=\frac{0+40.0}{2} \mathrm{~m} / \mathrm{s}=20.0 \mathrm{~m} / \mathrm{s}\) (c) \(x=v_{i x} t+\frac{1}{2} a t^{2}=0+\frac{1}{2}\left(8.00 \mathrm{~m} / \mathrm{s}^{2}\right)(5.00 \mathrm{~s})^{2}=100 \mathrm{~m} \quad\) or \(\quad x=v_{a v} t=(20.0 \mathrm{~m} / \mathrm{s})(5.00 \mathrm{~s})=100 \mathrm{~m}\)

A stone is thrown straight downward with initial speed \(8.0 \mathrm{~m} / \mathrm{s}\) from a height of \(25 \mathrm{~m}\). Find \((a)\) the time it takes to reach the ground and \((b)\) the speed with which it strikes.

A robot named Fred is initially moving at \(2.20 \mathrm{~m} / \mathrm{s}\) along a hallway in a space terminal. It subsequently speeds up to \(4.80 \mathrm{~m} / \mathrm{s}\) in a time of \(0.20 \mathrm{~s}\). Determine the size or magnitude of its average acceleration along the path traveled. The defining scalar equation is \(a_{a v}=\left(v_{f}-v_{i}\right) / t\). Everything is in proper SI units, so we need only carry out the calculation: $$ a_{a v}=\frac{4.80 \mathrm{~m} / \mathrm{s}-2.20 \mathrm{~m} / \mathrm{s}}{0.20 \mathrm{~s}}=13 \mathrm{~m} / \mathrm{s}^{2} $$ Notice that the answer has two significant figures because the time has only two significant figures.

A ball that is thrown vertically upward on the Moon returns to its starting point in \(4.0 \mathrm{~s}\). The acceleration due to gravity there is \(1.60\) \(\mathrm{m} / \mathrm{s}^{2}\) downward. Find the ball's original speed. Take up as positive. For the trip from beginning to end, \(y=0\) (it ends at the same level it started at), \(a=-1.60 \mathrm{~m} / \mathrm{s}^{2}, t=4.0 \mathrm{~s}\). Use \(y=v_{i y} t+\frac{1}{2} a t^{2}\) to find $$ 0=v_{i y}(4.0 \mathrm{~s})+\frac{1}{2}\left(-1.60 \mathrm{~m} / \mathrm{s}^{2}\right)(4.0 \mathrm{~s})^{2} $$ from which \(v_{i y}=3.2 \mathrm{~m} / \mathrm{s}\).

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