SATHEE: Motion in a Straight Line

Motion in a Straight Line - Result Question 17

17. A particle starts its motion from rest under the action of a constant force. If the distance covered in first 10 seconds is $S_{1}$ and that covered in the first 20 seconds is $S_{2}$ , then:

[2009]

(a) $S_{2} = 3 S_{1}$

(b) $S_{2} = 4 S_{1}$

(d) $S_{2} = 2 S_{1}$

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Answer:

Correct Answer: 17. (b)

Solution:

(b) $u = 0, t_{1} = 10 s, t_{2} = 20 s$

Using the relation, $S = u t + \frac{1}{2} a t^{2}$

Acceleration being the same in two cases,

$\begin{aligned} S_{1} = \frac{1}{2} a \times t_{1}^{2}, S_{2} = \frac{1}{2} a \times t_{2}^{2} \\ ∴ \frac{S_{1}}{S_{2}} = (\frac{t_{1}}{t_{2}})^{2} = (\frac{10}{20})^{2} = \frac{1}{4} \\ S_{2} = 4 S_{1} \end{aligned}$

Motion in a Straight Line - Result Question 17

17. A particle starts its motion from rest under the action of a constant force. If the distance covered in first 10 seconds is $S_{1}$ and that covered in the first 20 seconds is $S_{2}$ , then:

Answer:

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Motion in a Straight Line - Result Question 17

17. A particle starts its motion from rest under the action of a constant force. If the distance covered in first 10 seconds is S1 and that covered in the first 20 seconds is S2, then:

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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17. A particle starts its motion from rest under the action of a constant force. If the distance covered in first 10 seconds is $S_{1}$ and that covered in the first 20 seconds is $S_{2}$ , then: