| ⇦ | 
| ⇨ |
The distance travelled by a particle starting from rest and moving with an acceleration 4/3 ms⁻², in the third second is
Options
(a) 10/3 m
(b) 19/3 m
(c) 6 m
(d) 4 m
Correct Answer:
10/3 m
Explanation:
Distance travelled in the nth second is given by tₙ = u + a/2 (2n – 1)
put u= 0, a= 4/3 ms⁻², n= 3 d= 0 + 4/ 3×2 (2 x 3 – 1) = 4/6 x 5 = 10/ 3 m
Related Questions:
- The dimensional formula for Young’s modulus is
- The voltage gain of an amplifier with 9% negative feedback is 10.
- Which two of the following five physical parameters have the same dimensions?
- A point source is kept at a distnace 1000m has an illumination I. To change the illumination
- A body is moving in a vertical circle. The string must remain taut at the highest point
Topics: Motion in Straight Line
(93)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply