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Chapter 2: Problem 243

Displacement of a particle in y direction is given by \(\mathrm{y}=\mathrm{t}^{2}-5 \mathrm{t}+5\) where \(\mathrm{t}\) is in second. Calculate the time when its velocity is zero. (A) \(5 \mathrm{~s}\) (B) \(2.5 \mathrm{~s}\) (C) \(10 \mathrm{~s}\) (D) \(3 \mathrm{~s}\)

Short Answer

Expert verified
The correct answer is (B) \(2.5~s\).

Step by step solution

01

Identify the given displacement function

The displacement function in the y-direction is given by: \[y(t) = t^2 - 5t + 5\]
02

Differentiate the displacement function

To find the velocity function, we need to differentiate the displacement function y(t) with respect to time t. Using basic differentiation rules, we get: \( \frac{d}{dt}(t^2 - 5t + 5) = \frac{d}{dt}(t^2) - \frac{d}{dt}(5t) + \frac{d}{dt}(5)\) Applying power rule and constant rule in differentiation: \(v(t) = 2t - 5\)
03

Find when the velocity is zero

Now, we need to find the time when the velocity function v(t) is zero. So, we set v(t) equal to zero and solve for t: \(0 = 2t - 5\)
04

Solve for t

To solve the equation \(0 = 2t - 5\), add 5 to both sides of the equation and then divide by 2: \[t = \frac{5}{2} = 2.5\] So, the time when the velocity is zero is 2.5 seconds. The correct answer is (B) \(2.5~s\).

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Most popular questions from this chapter

\(x\) and y co-ordinates of a particle moving in \(\mathrm{x}-\mathrm{y}\) plane at some instant are \(\mathrm{x}=2 \mathrm{t}^{2}\) and $\mathrm{y}=(3 / 2) \mathrm{t}^{2}\( Calculate y co-ordinate when its \)\mathrm{x}\( coordinate is \)8 \mathrm{~m}$. (A) \(3 \mathrm{~m}\) (B) \(6 \mathrm{~m}\) (C) \(8 \mathrm{~m}\) (D) \(9 \mathrm{~m}\)

Find a unit vector in direction of \(\hat{1}+2 \hat{\jmath}-3 \mathrm{k}\) (A) \((1 / \sqrt{7})(\hat{1}+2 \hat{\jmath}-3 \mathrm{k})\) (B) \(-(1 / 2)(\hat{1}+2 \hat{\jmath}-3 \mathrm{k})\) (C) \((1 / \sqrt{14})(\hat{1}+2 \hat{j}-3 \mathrm{k})\) (D) \((1 / \sqrt{5})(\hat{1}+2 \hat{\jmath}-3 \mathrm{k})\)

A particle is thrown in upward direction with Velocity \(\mathrm{V}_{0}\). It passes through a point \(\mathrm{p}\) of height \(\mathrm{h}\) at time \(\mathrm{t}_{1}\) and \(\mathrm{t}_{2}\) so \(\mathrm{t}_{1}+\mathrm{t}_{2}=\ldots\) (A) \(\left(\mathrm{v}_{0} / \mathrm{g}\right)\) (B) \(\left[\left(2 \mathrm{v}_{0}\right) / \mathrm{g}\right]\) (C) \((2 \mathrm{~h} / \mathrm{g})\) (D) \((\mathrm{h} / 2 \mathrm{~g})\)

Comprehensions type questions. A particle is moving in a circle of radius \(\mathrm{R}\) with constant speed. The time period of the particle is T Now after time \(\mathrm{t}=(\mathrm{T} / 6)\) Average speed of the particle is (A) \((\pi \mathrm{R} / 6 \mathrm{~T})\) (B) \([(2 \pi R) / 3 \mathrm{~T}]\) (C) \([(2 \pi R) / T]\) (D) \((\mathrm{R} / \mathrm{T})\)

The resultant of two forces of magnitude \(2 \mathrm{~N}\) and \(3 \mathrm{~N}\) can never be. (A) \(4 \mathrm{~N}\) (B) \(1 \mathrm{~N}\) (C) \(2.5 \mathrm{~N}\) (D) \((1 / 2) \mathrm{N}\)
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