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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 matching pairs are: (A) and (1), (B) and (2), (C) and (3) and (D) and (4). Therefore, the distance travelled in first 9s is 63m, the magnitude of acceleration is 2 m/s虏, the displacement in first 7s is 65m and the time when displacement is zero is 16s.

Step by step solution

01

Understand the Problem

A particle moves in a straight line with uniform acceleration and displaces by 13 m in the second second and 7 m in the fifth second. It is necessary to find the distance travelled in the first 9 seconds, the magnitude of acceleration, the displacement in the first 7 seconds, and the time when the displacement is zero.
02

Calculation of Acceleration

The displacement in nth second \(s_n\) can be written as, \(s_n = u + (n - 1/2) a\), where u is the initial velocity and a is the acceleration. The displacement in the second second is 13 m, and that in the fifth second is 7 m. Solving these two equations (second second and fifth second), we get acceleration a as 2 m/s虏.
03

Calculation of Initial Velocity

Substitute the acceleration value obtained in previous step in the equation for the second second, we get initial velocity u as 11 m/s.
04

Distance Travelled in First 9 Seconds

Use the first equation of motion \(s = ut + 1/2 a t^2\), where s is the distance and t is time. By substituting u = 11 m/s, a = 2 m/s虏, and t = 9 s, we get s = 63 m.
05

Displacement in First 7 Seconds

Again, use the first equation of motion and by substituting u, a and t = 7 s, we get s = 65 m.
06

Time when Displacement is Zero

The displacement becomes zero when the particle comes to rest. Use the equation \(v = u + at\), where v is the final velocity and substituting v = 0, u = 11 m/s, a = 2 m/s虏, we get t = 16 s.

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Most popular questions from this chapter

A particle is moving eastwards with a velocity of \(4 \mathrm{~m} / \mathrm{s}\). In \(5 \mathrm{~s}\) the velocity changes to \(3 \mathrm{~m} / \mathrm{s}\) northwards. The average acceleration in this time interval is (A) \(\frac{1}{2} \mathrm{~m} / \mathrm{s}^{2}\) towards north-east (B) \(1 \mathrm{~m} / \mathrm{s}^{2}\) towards north-west (C) \(\frac{1}{\sqrt{2}} \mathrm{~m} / \mathrm{s}^{2}\) towards north-east (D) \(\frac{1}{2} \mathrm{~m} / \mathrm{s}^{2}\) towards north-west

The velocity of a particle is \(v=v_{0}+g t+f t^{2}\). If its position is \(x=0\) at \(t=0\), then its displacement after unit time \((t=1)\) is (A) \(v_{0}+g / 2+f\) (B) \(v_{0}+2 g+3 f\) (C) \(v_{0}+g / 2+f / 3\) (D) \(v_{0}+g+f\)

Two trains \(\mathrm{A}\) and \(\mathrm{B}\) of length \(400 \mathrm{~m}\) each are moving on two parallel tracks with a uniform speed of \(72 \mathrm{~km} \mathrm{~h}^{-1}\) in the same direction, with \(\mathrm{A}\) ahead of \(\mathrm{B}\). The driver of B decides to overtake A and accelerates by \(1 \mathrm{~m} / \mathrm{s}^{2}\). If after \(50 \mathrm{~s}\), the guard of B just brushes past the driver of \(\mathrm{A}\), what was the original distance between them? (A) \(2250 \mathrm{~m}\) (B) \(1250 \mathrm{~m}\) (C) \(1000 \mathrm{~m}\) (D) \(2000 \mathrm{~m}\)

The position \((x)\) of a particle varies with time as \(x=\) \(t^{3}-3 t^{2}\), where \(t\) is time in second. Match Column-I with Column-II. $$ \begin{array}{ll} \hline \text { Column-I } & \text { Column-II } \\ \hline \text { (A) Position of particle is } & \text { (1) } t=3 \mathrm{~s} \\\ \text { zero at } \\ \text { (B) Velocity of particle is } & \text { (2) } t=2 \mathrm{~s} \\ \text { zero at } \\ \text { (C) Magnitude of } & \text { (3) } t=5 \mathrm{~s} \\ \text { acceleration of particle } \\ \text { is zero at } \\ \begin{array}{ll} \text { (D) Magnitude of } \\ \text { instantaneous } \\ \text { acceleration is equal to } \\ 18 \mathrm{~m} / \mathrm{s}^{2} \end{array} & \text { (4) } t=1 \mathrm{~s} \\ \hline \end{array} $$

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
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