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Chapter 2: Problem 65

A particle moves along the parabolic path \(y=a x^{2}\) in such a way that the \(x\) component of the velocity remains constant, say \(c\). The acceleration of the particle is (A) \(a c \hat{k}\) (B) \(2 a c^{2} \hat{j}\) (C) \(2 a c^{2} \hat{k}\) (D) \(a^{2} c \hat{j}\)

Short Answer

Expert verified
The acceleration of the particle is found by combining the x and y components of the acceleration. The x-component of the acceleration is zero since the x-component of the velocity is constant. The y-component of the acceleration can be found by differentiating the parabolic path \(y = ax^2\) with respect to time twice. The final acceleration is \(2ac^2\hat{j}\), which corresponds to option (B).

Step by step solution

01

Find the x-component of the acceleration

Since the x-component of velocity (c) is constant, the x-component of acceleration would be zero. Using the given equation for the path, \(y = ax^2\), we will calculate the y-component of the acceleration.
02

Differentiate the path equation w.r.t time to get the y-component of the velocity

To find the y-component of the velocity, we should differentiate the given path equation \(y = ax^2\) with respect to time. First, let's differentiate it with respect to x using the chain rule: \(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\) Since we know that the x-component of the velocity is constant and equal to \(c\), \(\frac{dx}{dt} = c\). Now differentiate \(y = ax^2\) with respect to x: \(\frac{dy}{dx} = 2ax\) Now, multiply \(\frac{dy}{dx}\) by \(\frac{dx}{dt}\) (which is c): \(\frac{dy}{dt} = 2ax \cdot c\) The y-component of the velocity is then \(2acx\).
03

Differentiate the y-component of the velocity w.r.t time to get the y-component of the acceleration

Now, differentiate the y-component of the velocity w.r.t time to get the y-component of the acceleration: \(\frac{d^2y}{dt^2} = \frac{d(2acx)}{dt}\) Using the chain rule again: \(\frac{d^2y}{dt^2} = \frac{d(2acx)}{dx} \cdot \frac{dx}{dt}\) Differentiate \(2acx\) w.r.t x: \(\frac{d(2acx)}{dx} = 2ac\) Now, multiply \(\frac{d(2acx)}{dx}\) by \(\frac{dx}{dt}\) (which is c): \(\frac{d^2y}{dt^2} = 2ac \cdot c\) The y-component of the acceleration is then \(2ac^2\).
04

Combine the x and y components of the acceleration and choose the correct option

Since the x-component of the acceleration is zero and the y-component is \(2ac^2\), the acceleration of the particle is \(2ac^2\hat{j}\). This corresponds to option (B) \(2ac^2\hat{j}\). So the correct answer is (B).

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Most popular questions from this chapter

The position of a particle is given by $$ r=3.0 t \hat{i}-2.0 t^{2} \hat{j}+4.0 \hat{k} \mathrm{~m} $$ Where \(t\) is in seconds and the coefficients have the proper units for \(r\) to be in metres. What is the magnitude of velocity of the particle \(t=2.0 \mathrm{~s}\) ? (A) \(\sqrt{72} \mathrm{~m} / \mathrm{s}\) (B) \(\sqrt{41} \mathrm{~m} / \mathrm{s}\) (C) \(\sqrt{11} \mathrm{~m} / \mathrm{s}\) (D) None

A projectile is thrown horizontally from top of a building of height \(10 \mathrm{~m}\) with certain speed \((u)\). At the same time another projectile is thrown from ground \(10 \mathrm{~m}\) away from the building with equal speed \((u)\) on the same vertical plane. If they collide after \(2 s\), then choose the correct options. (A) The angle of projection for second projectile is \(60^{\circ}\) and \(u=10 \mathrm{~ms}^{-1}\) (B) The angle of projection for second projectile is \(90^{\circ}\) and \(u=5 \mathrm{~ms}^{-1}\) (C) The angle of projection for second projectile is \(60^{\circ}\) and \(u=5 \mathrm{~ms}^{-1}\) (D) The angle of projection for second projectile is \(45^{\circ}\) and \(u=10 \mathrm{~ms}^{-1}\)

A glass wind screen whose inclination with the vertical can be changed is mounted on a car. The car moves horizontally with a speed of \(2 \mathrm{~m} / \mathrm{s}\). If the angle of the wind screen with vertical is \(\theta\) when vertically downward falling raindrops with velocity of \(6 \mathrm{~m} / \mathrm{s}\) strikes the screen perpendicularly. Find \(\tan \theta\).

The maximum height attained by a projectile is increased by \(5 \%\), keeping the angle of projection constant. The corresponding percentage increase in horizontal range will be (A) \(5 \%\) (B) \(10 \%\) (C) \(15 \%\) (D) \(20 \%\)

A particle is projected with a speed of \(40 \mathrm{~m} / \mathrm{s}\) at an angle of \(60^{\circ}\) with the horizontal. At what height speed of particle becomes half of initial speed \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\). (A) \(30 \mathrm{~m}\) (B) \(45 \mathrm{~m}\) (C) \(37.5 \mathrm{~m}\) (D) \(60 \mathrm{~m}\)
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