Question
Question: A particle moves from the point \(\left( {2.0\hat i + 4.0\hat j} \right)m\) at \(t = 0\), with an in...
A particle moves from the point (2.0i^+4.0j^)m at t=0, with an initial velocity (5.0i^+4.0j^)ms−1. It is acted upon by a constant force which produces a constant acceleration (4.0i^+4.0j^)ms−2. What is the distance of the particle from the origin at time 2s ?
A. 202m
B. 102m
C. 5m
D. 15m
Solution
To solve the question we will use the second equation of motion in vector form s=ut+21at2 to and then equate with difference of final and initial position will give the distance and then find magnitude by ∣A∣=Ax2+Ay2+Az2.
Complete step by step answer:
Now from the question
Given initial position vector ri=(2.0i^+4.0j^)m
Initial velocity vector u=(5.0i^+4.0)ms−1
Acceleration vector a=(4.0i^+4.0j^)ms−2
Now from second equation of motion we have
s=ut+21at2
Now s=(5.0i^+4.0j^)(2)+21(4.0i^+4.0j^)(2)2
s=(18i^+16j^)
Now the distance covered by particle from origin at 2s