ACCELERATION \[a=\frac{6-0}{1}=6m{{s}^{-2}}\] For \[t=0\] to \[t=1\]s, \[{{S}_{1}}=\frac{1}{2}\times 6{{(1)}^{2}}=3m\] (i) For \[t=1\] s to \[t=2\] s, \[{{S}_{2}}=6.1-\frac{1}{2}\times 6{{(1)}^{2}}=3m\] (ii) For \[t=2s\] to \[t=3s\], \[{{S}_{3}}=0-\frac{1}{2}\times 6{{(1)}^{2}}=-3m\] (iii) Total DISPLACEMENT \[\text{S=}{{\text{S}}_{\text{1}}}\text{+}{{\text{S}}_{\text{2}}}\text{+}{{\text{S}}_{\text{3}}}\text{=3m}\] Average velocity\[\text{=}\frac{\text{3}}{\text{3}}\text{=1M}{{\text{s}}^{\text{-1}}}\] Total DISTANCE travelled \[\text{=9m}\] Average speed\[\text{=}\frac{9}{3}=3m{{s}^{-1}}\]

"> ACCELERATION \[a=\frac{6-0}{1}=6m{{s}^{-2}}\] For \[t=0\] to \[t=1\]s, \[{{S}_{1}}=\frac{1}{2}\times 6{{(1)}^{2}}=3m\] (i) For \[t=1\] s to \[t=2\] s, \[{{S}_{2}}=6.1-\frac{1}{2}\times 6{{(1)}^{2}}=3m\] (ii) For \[t=2s\] to \[t=3s\], \[{{S}_{3}}=0-\frac{1}{2}\times 6{{(1)}^{2}}=-3m\] (iii) Total DISPLACEMENT \[\text{S=}{{\text{S}}_{\text{1}}}\text{+}{{\text{S}}_{\text{2}}}\text{+}{{\text{S}}_{\text{3}}}\text{=3m}\] Average velocity\[\text{=}\frac{\text{3}}{\text{3}}\text{=1M}{{\text{s}}^{\text{-1}}}\] Total DISTANCE travelled \[\text{=9m}\] Average speed\[\text{=}\frac{9}{3}=3m{{s}^{-1}}\]

">

A toy car with charge q moves on a frictionless horizontal plane surface under the influence of a uniform electric field\[\overrightarrow{\text{E}}\]. Due to the force q\[\overrightarrow{\text{E}}\], its velocity increases from 0 to 6 m/s in one second duration. At that instant the direction of the field is reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between 0 to 3 seconds are respectively [NEET - 2018]

NEET Physics in NEET 10 months ago

  7   0   0   0   0 tuteeHUB earn credit +10 pts

5 Star Rating 1 Rating

[b] \[\text{v=-6m}{{\text{s}}^{-1}}\] ACCELERATION \[a=\frac{6-0}{1}=6m{{s}^{-2}}\] For \[t=0\] to \[t=1\]s, \[{{S}_{1}}=\frac{1}{2}\times 6{{(1)}^{2}}=3m\] (i) For \[t=1\] s to \[t=2\] s, \[{{S}_{2}}=6.1-\frac{1}{2}\times 6{{(1)}^{2}}=3m\] (ii) For \[t=2s\] to \[t=3s\], \[{{S}_{3}}=0-\frac{1}{2}\times 6{{(1)}^{2}}=-3m\] (iii) Total DISPLACEMENT \[\text{S=}{{\text{S}}_{\text{1}}}\text{+}{{\text{S}}_{\text{2}}}\text{+}{{\text{S}}_{\text{3}}}\text{=3m}\] Average velocity\[\text{=}\frac{\text{3}}{\text{3}}\text{=1M}{{\text{s}}^{\text{-1}}}\] Total DISTANCE travelled \[\text{=9m}\] Average speed\[\text{=}\frac{9}{3}=3m{{s}^{-1}}\]

Posted on 25 Aug 2024, this text provides information on NEET related to Physics in NEET. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

Take Quiz To Earn Credits!

Turn Your Knowledge into Earnings.

tuteehub_quiz

Tuteehub forum answer Answers

Post Answer

No matter what stage you're at in your education or career, TuteeHub will help you reach the next level that you're aiming for. Simply,Choose a subject/topic and get started in self-paced practice sessions to improve your knowledge and scores.