A particle moves from the point (2.0 hati + 4.0 hatj) m, at t = 0, with an initial velocity (5.0 hati + 4.0 hatj) ms-1 . It is acted upon by a constant force which produces a constant acceleration (4.0 hati + 4.0 hatj) ms-2. What is the distance of the particle from the origin at time 2 s ?

Question

A particle moves from the point (2.0 hati + 4.0 hatj) m, at t = 0, with an initial velocity (5.0 hati + 4.0 hatj) ms-1 . It is acted upon by a constant force which produces a constant acceleration (4.0 hati + 4.0 hatj) ms-2. What is the distance of the particle from the origin at time 2 s ? (A) 20 √2 m (B) 10 √2 m (C) 5 m (D) 15 m

Tardigrade Admin · Accepted Answer

vecS = (5 hati +4 hatj)2+ (1/2) (4 hati +4 hatj)4 = 10 hati + 8 hatj +8 hati + 8 hatj vecrf - vecri = 18 hati+16 hatj vecrf = 20 hati + 20 hatj | vecrf| = 20 √2

$202 m$

$102 m$

$5 m$

$15 m$

$S = (5 \hat{i} + 4 \hat{j}) 2 + \frac{1}{2} (4 \hat{i} + 4 \hat{j}) 4$
$= 10 \hat{i} + 8 \hat{j} + 8 \hat{i} + 8 \hat{j}$
$r_{f} - r_{i} = 18 \hat{i} + 16 \hat{j}$
$r_{f} = 20 \hat{i} + 20 \hat{j}$
$∣ r_{f} ∣ = 202$

Q. 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 ?

202​m

102​m

5m

15m

S=(5i^+4j^​)2+21​(4i^+4j^​)4 =10i^+8j^​+8i^+8j^​ rf​−ri​=18i^+16j^​ rf​=20i^+20j^​ ∣rf​∣=202​

$202 m$

$102 m$

$5 m$

$15 m$

$S = (5 \hat{i} + 4 \hat{j}) 2 + \frac{1}{2} (4 \hat{i} + 4 \hat{j}) 4$
$= 10 \hat{i} + 8 \hat{j} + 8 \hat{i} + 8 \hat{j}$
$r_{f} - r_{i} = 18 \hat{i} + 16 \hat{j}$
$r_{f} = 20 \hat{i} + 20 \hat{j}$
$∣ r_{f} ∣ = 202$