Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Starting from the origin at time t=0, with initial velocity , a particle moves in the x-y plane with constant accelertaion of <img

Starting from the origin at time t=0, with initial velocity 5\widehat{j}ms^{-1}, a particle moves in the x-y plane with constant accelertaion of \left (10\widehat{i} +4\widehat{j} \right )ms^{-2}. At time t, its coordinates are (20m,y_{0}\ m). The values of t and y_{0} are , respectively:
Option: 1 2 s and 18 m
Option: 2 4 s and 52 m
Option: 3 5 s and 24 m
Option: 4 5 s and 25 m

Answers (1)

best_answer

Given

\vec{a}=\left (10\widehat{i} +4\widehat{j} \right )ms^{-2}

 \vec{u}=5\widehat{j}ms^{-1}

\text { And final coordinates }\left(20, y_{0}\right) \text { in time } t

So

For x-axis

\begin{array}{l} \mathrm{S}_{\mathrm{x}}=u_{\mathrm{x}} \mathrm{t}+\frac{1}{2} \mathrm{a}_{\mathrm{x}} \mathrm{t}^{2} \\ 20-0=0+(\frac{1}{2} \times 10 \times \mathrm{t}^{2}) \\ \mathrm{t}=2 \mathrm{sec} \end{array}

For y-axis

\begin{array}{l} \mathrm{S}_{\mathrm{y}}=\mathrm{u}_{\mathrm{y}} \times \mathrm{t}+\frac{1}{2} \mathrm{a}_{\mathrm{y}} \mathrm{t}^{2} \\ \mathrm{y}_{0}=(5 \times 2)+(\frac{1}{2} \times4 \times 2^{2})=18 \mathrm{~m} \end{array}

So answer is 2 s and 18 m

Posted by

avinash.dongre

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

Amrita BTech Admission 2025: JEE Main-based registration deadline extended to August 30

Latest Question