/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 168 A particle moving along a straig... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 168

A particle moving along a straight line with uniform acceleration displaces by \(13 \mathrm{~m}\) and \(7 \mathrm{~m}\) in second and fifth second of its motion respectively. Column-I (A) Distance (in \(\mathrm{m}\) ) travelled in first \(9 \mathrm{~s}\) (B) Magnitude of acceleration (in \(\mathrm{ms}^{-2}\) ) of particle (C) Displacement (in \(\mathrm{m}\) ) in first \(7 \mathrm{~s}\) (D) Time (in s) when displacement is zero Column-II (1) 63 (2) 2 (3) 65 (4) 16

Short Answer

Expert verified
The short answer is: (B) Magnitude of acceleration 鈫� \(2 \: \mathrm{m/s^2}\) (2)

Step by step solution

01

List the given information

The particle displaces by 13 m in the second second and 7 m in the fifth second of its motion.
02

Calculate the initial velocity and acceleration

We know the formulas for displacement in the nth second for a particle have uniform acceleration as: \(S_n = u + (n - 1) \times a\) Let's use the formula for the second and the fifth seconds. For the second second (n = 2): \(13 = u + a\) For the fifth second (n = 5): \(7 = u + 4a\) Now we have a system of equations with two unknowns, u (initial velocity) and a (acceleration). To solve this system, let's subtract the first equation from the second equation: \(-6 = 3a\) \( a = -2 \: \text{m/s}^2 \) Now substitute the value of a in the first equation and find the initial velocity: \(13 = u - 2\) \( u = 15 \: \text{m/s} \) The initial velocity is 15 m/s, and the acceleration is -2 m/s虏.
03

Find the distance traveled in the first 9 seconds

To find the distance traveled in the first 9 seconds, we can use the equation of motion: \(S = ut + \frac{1}{2}at^2\) Plug in the values of u, a, and t (t = 9): \(S = 15 \times 9 + \frac{1}{2}(-2)(9^2)\) \( S = 135 - 81 = 54 \: \text{m} \) So the distance traveled in the first 9 seconds (A) is 54 m.
04

Find the magnitude of acceleration (B) of the particle

We have already calculated the acceleration of the particle in Step 2: The magnitude of acceleration (B) is: \( |a| = |-2| = 2\: \text{m/s}^2 \)
05

Calculate displacement in the first 7 seconds (C)

To find the displacement in the first 7 seconds, we will use the equation of motion again: \(S = ut + \frac{1}{2}at^2\) Plug in the values of u, a, and t (t = 7): \(S = 15 \times 7 + \frac{1}{2}(-2)(7^2)\) \( S = 105 - 49 = 56 \: \text{m} \) So displacement (C) in the first 7 seconds is 56 m.
06

Calculate the time when displacement is zero (D)

To find the time when displacement is zero, we will use the equation of motion: \(S = ut + \frac{1}{2}at^2\) Set S to 0 and plug in the values of u and a: \(0 = 15t - t^2\) Factor t out of the equation: \(t(15 - t) = 0\) The solutions for t are: \(t = 0 \: \text{s}\) (initial condition) \( t = 15 \: \text{s} \) So the time (D) when displacement is zero is 15 s. Column-I (A) Distance (in \(\mathrm{m}\) ) traveled in first \(9 \mathrm{~s}\) 鈫� 54 (Not given) (B) Magnitude of acceleration (in \(\mathrm{ms}^{-2}\) ) of particle 鈫� 2 (2) (C) Displacement (in \(\mathrm{m}\) ) in first \(7 \mathrm{~s}\) 鈫� 56 (Not given) (D) Time (in s) when displacement is zero 鈫� 15 (Not given) Column-II (1) 63 is not connected to any option in Column-I (3) 65 is not connected to any option in Column-I (4) 16 is not connected to any option in Column-I

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Most popular questions from this chapter

A coin is dropped in a lift. It takes time \(t_{1}\) to reach the floor when lift is stationary. It takes time \(t_{2}\) when lift is moving up with constant acceleration, then (A) \(t_{1}>t_{2}\) (B) \(t_{2}>t_{1}\) (C) \(t_{1}=t_{2}\) (D) \(t_{1} \gg t_{2}\)

A particle is thrown with a speed of \(12 \mathrm{~m} / \mathrm{s}\) at an angle \(60^{\circ}\) with the horizontal. The time interval between the moments when its speed is \(10 \mathrm{~m} / \mathrm{s}\) is \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (A) \(1.0 \mathrm{~s}\) (B) \(1.2 \mathrm{~s}\) (C) \(1.4 \mathrm{~s}\) (D) \(1.6 \mathrm{~s}\)

One body is dropped, while a second body is thrown downward with an initial velocity of \(1 \mathrm{~ms}^{-1}\) simultaneously. The separation between these is \(1.8 \mathrm{~m}\) after a time (A) \(4.5 \mathrm{~s}\) (B) \(9 \mathrm{~s}\) (C) \(1.8 \mathrm{~s}\) (D) \(36 \mathrm{~s}\)

A particle has an initial velocity \(11 \mathrm{~m} / \mathrm{s}\) due east and a constant acceleration of \(2 \mathrm{~m} / \mathrm{s}^{2}\) due west. The distance covered by the particle in sixth second is (A) Zero (B) \(0.5 \mathrm{~m}\) (C) \(1 \mathrm{~m}\) (D) \(2 \mathrm{~m}\)

A \(2 \mathrm{~m}\) wide truck is moving with a uniform speed \(v_{0}=\) \(8 \mathrm{~m} / \mathrm{s}\) along a straight horizontal road. A pedestrian starts to cross the road with a uniform speed \(v\) when the truck is \(4 \mathrm{~m}\) away from him. The minimum value of \(v\) so that he can cross the road safely is (A) \(2.62 \mathrm{~m} / \mathrm{s}\) (B) \(4.6 \mathrm{~m} / \mathrm{s}\) (C) \(3.57 \mathrm{~m} / \mathrm{s}\) (D) \(1.414 \mathrm{~m} / \mathrm{s}\)
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