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Chapter 5: Problem 13

Aparticle has mitial velocity \(4 i+4 j \mathrm{~m} \mathrm{~s}^{-1}\) and an acceleration \- \(0.4 i \mathrm{~ms}^{-2}\), at what time will its speed be \(5 \mathrm{~ms}^{-1} ?\) (1) \(2.5 \mathrm{~s}\) (2) \(17.5 \mathrm{~s}\) (3) s (4) \(8.5 \mathrm{~s}\)

Short Answer

Expert verified
The particle will have a speed of 5 m/s at 2.5 seconds.

Step by step solution

01

Understanding the Problem

We are given the initial velocity \( \vec{u} = 4i + 4j \text{ m/s} \) and an acceleration \( \vec{a} = 0.4i \text{ m/s}^2 \). We need to find the time when the particle's speed becomes \( 5 \text{ m/s} \).
02

Express Velocity as a Function of Time

The velocity \( \vec{v} \) of the particle as a function of time \( t \) can be expressed as \( \vec{v}(t) = \vec{u} + \vec{a}t \). Substituting the given values, we get: \( \vec{v}(t) = (4 + 0.4t)i + 4j \).
03

Calculate the Speed Expression

The speed \( v \) is the magnitude of the velocity vector \( \vec{v}(t) \): \( v = \sqrt{(4 + 0.4t)^2 + 4^2} \). Simplifying gives \( v = \sqrt{(4 + 0.4t)^2 + 16} \).
04

Set the Expression Equal to the Desired Speed

We need the speed to be \( 5 \text{ m/s} \). Set the expression for speed equal to 5: \( \sqrt{(4 + 0.4t)^2 + 16} = 5 \).
05

Square Both Sides of the Equation

Square both sides to remove the square root: \( (4 + 0.4t)^2 + 16 = 25 \). Simplify to get \( (4 + 0.4t)^2 = 9 \).
06

Solve for Time t

Solve \( (4 + 0.4t)^2 = 9 \). Take the square root of both sides: \( 4 + 0.4t = 3 \) or \( 4 + 0.4t = -3 \). Solving these gives, \( 0.4t = -1 \) or \( 0.4t = -7 \). Thus, \( t = 2.5 \text{ s} \) or \( t = -17.5 \text{ s} \). Since time cannot be negative, \( t = 2.5 \text{ s} \).

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Key Concepts

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

Initial Velocity
The initial velocity of a projectile is an important concept when studying its motion. It's essentially the velocity that the particle starts with the moment it's released. In vector notation, this can look like something derived from combining different directional components, like in this scenario.
  • In our problem, the initial velocity is given as \( \vec{u} = 4i + 4j \text{ m/s} \), meaning the particle is initially moving at 4 meters per second in the x-direction (along the i-axis) and 4 meters per second in the y-direction (along the j-axis).
  • Understanding initial velocity is crucial because it sets the starting conditions for predicting how the object will move over time.
This combined vector notation provides a holistic understanding of the particle's starting speed and direction in a two-dimensional plane.
Acceleration
Acceleration is another key concept in physics, and it dictates how the velocity of the particle changes over time. It's essentially the rate at which the object's speed or direction changes.
  • In our specific problem, acceleration \( \vec{a} = 0.4i \text{ m/s}^2 \) indicates that the particle is experiencing an acceleration solely in the x-direction.
  • This means each second, the velocity of the particle increases by 0.4 meters per second in the i-direction. There is no change in the j-direction, so the vertical component of acceleration is 0.
Acceleration helps us forecast how the particle's motion evolves, allowing us to compute the velocity or position at a future time.
Velocity as a Function of Time
To anticipate how a particle moves, it's crucial to express its velocity as a function of time. This tells us how speed and direction evolve as time passes. The formula to express velocity is:
\[\vec{v}(t) = \vec{u} + \vec{a}t\]Using the given initial velocity and acceleration, we derive:
  • \( \vec{v}(t) = (4 + 0.4t)i + 4j \), where 4 m/s in the j-direction remains constant due to no vertical acceleration.
This equation is essential as it determines the velocity of the particle at any given time, accounting for changes due to acceleration.
Magnitude of Velocity
The magnitude of velocity is what we commonly refer to as speed. It's the overall rate of movement regardless of direction. For a two-dimensional velocity vector, the magnitude (or speed) can be computed using:
\[v = \sqrt{(v_x)^2 + (v_y)^2}\]Substituting our expression for velocity:
  • \( v = \sqrt{(4 + 0.4t)^2 + 4^2} \)
Setting this to the desired speed, say 5 m/s, allows us to solve for the time when the particle achieves this speed. It encapsulates how fast the particle is actually moving, not just in one direction but considering its overall movement.

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Most popular questions from this chapter

Rain appears to fall vertically to a man walking at \(3 \mathrm{~km} \mathrm{~h}^{-1}\), but when he changes his speed to double, the rain appears to fall at \(45^{\circ}\) with vertical. Study the following statements and find which of them are correct. i. Velocity of rain is \(2 \sqrt{3} \mathrm{~km} \mathrm{~h}^{-1}\) ii. The angle of fall of rain (with vertical) is $$ \theta=\tan ^{-1}\left(\frac{1}{\sqrt{2}}\right) \text {. } $$ iii. The angle of fall of rain (with vertical) is $$ \theta=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right) \text {. } $$ iv. Velocity of rain is \(3 \sqrt{2} \mathrm{kmh}^{-1}\). (I) Statements (i) and (ii) are correct. (2) Statements (i) and (iii) are correct. (3) Statements (iii) and (iv) are correct. (4) Statements (ii) and (iv) are correct.

A helicopter is flying at \(200 \mathrm{~m}\) and flying at \(25 \mathrm{~m} \mathrm{~s}^{-1}\) at an angle \(37^{\circ}\) above the horizontal when a package is dropped from it. If the helicopter flies at constant velocity, find the \(x\) and coordinates of the location of the helicopter when the \(v\) package lands. (1) \(160 \mathrm{~m}, 320 \mathrm{~m}\) (2) \(100 \mathrm{~m}, 200 \mathrm{~m}\) (3) \(200 \mathrm{~m}, 400 \mathrm{~m}\) (4) \(50 \mathrm{~m}, 100 \mathrm{~m}\)

Jai is standing on the top of a building of height \(25 \mathrm{~m}\) he wants to throw his gun to Veeru who stands on top of another building of height \(20 \mathrm{~m}\) at distance \(15 \mathrm{~m}\) from first building. For which horizontal speed of projection, it is possible? (1) \(5 \mathrm{~ms}^{-1}\) (2) \(10 \mathrm{~ms}^{-1}\) (3) \(15 \mathrm{~ms}^{-1}\) (4) \(20 \mathrm{~ms}^{-1}\)

equation of motion of a projectile is \(y=12 x-\frac{3}{4} x^{2}\). The horizontal component of velocity is \(3 \mathrm{~ms}^{-1}\). What is The hortzo of the projectile? (1) \(18 \mathrm{~m}\) (4) \(21.6 \mathrm{~m}\) (3) \(12 \mathrm{~m}\)

A helicopter is flying at \(200 \mathrm{~m}\) and flying at \(25 \mathrm{~m} \mathrm{~s}^{-1}\) at an angle \(37^{\circ}\) above the horizontal when a package is dropped from it. The distance of the point from point \(O\) where the package lands is (1) \(80 \mathrm{~m}\) (2) \(100 \mathrm{~m}\) (3) \(200 \mathrm{~m}\) (4) \(160 \mathrm{~m}\)
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