A particle leaves the origin with an initial velocity overset arrow textv = ( text3.00 hat texti) ( textms)- 1 and a constant acceleration overset arrow texta = (- text1.00 hat texti - text0.5 hat textj) ( textms)- 2 . When the particle reaches it maximum x -coordinate, what is the magnitude of its velocity (in m/s) in y -direction?

Question

A particle leaves the origin with an initial velocity overset arrow textv = ( text3.00 hat texti) ( textms)- 1 and a constant acceleration overset arrow texta = (- text1.00 hat texti - text0.5 hat textj) ( textms)- 2 . When the particle reaches it maximum x -coordinate, what is the magnitude of its velocity (in m/s) in y -direction? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Velocity along x direction is vx = v0 2 + textitat vx = 3 + (- 1) t so when particle reach maximum x coordinate vx = 0 ⇒ t = 3 sec similarly vy = v0 y + ay t vy = 0 + (- (1/2)) t (vy)at t = 3=-(3/2)=-1.5 m s- 1

Q. A particle leaves the origin with an initial velocity $v \to = (3.00 \hat{i}) (ms)^{- 1}$ and a constant acceleration $a \to = (- 1.00 \hat{i} - 0.5 \hat{j}) (ms)^{- 2}$ . When the particle reaches it maximum $x$ -coordinate, what is the magnitude of its velocity (in m/s) in $y$ -direction?

Velocity along $x$ direction is
$v_{x} = v_{02} + at$
$v_{x} = 3 + (- 1) t$
so when particle reach maximum $x$ coordinate $v_{x} = 0 \Rightarrow t = 3 sec$
similarly $v_{y} = v_{0 y} + a_{y} t$
$v_{y} = 0 + (- \frac{1}{2}) t$
$(v_{y})_{a t t = 3} = - \frac{3}{2} = - 1.5 m s^{- 1}$

Q. A particle leaves the origin with an initial velocity v→=(3.00i^)(ms)−1 and a constant acceleration a→=(−1.00i^−0.5j^​)(ms)−2 . When the particle reaches it maximum x -coordinate, what is the magnitude of its velocity (in m/s) in y -direction?

Velocity along x direction is vx​=v02​+at vx​=3+(−1)t so when particle reach maximum x coordinate vx​=0⇒t=3sec similarly vy​=v0y​+ay​t vy​=0+(−21​)t (vy​)att=3​=−23​=−1.5ms−1

Q. A particle leaves the origin with an initial velocity $v \to = (3.00 \hat{i}) (ms)^{- 1}$ and a constant acceleration $a \to = (- 1.00 \hat{i} - 0.5 \hat{j}) (ms)^{- 2}$ . When the particle reaches it maximum $x$ -coordinate, what is the magnitude of its velocity (in m/s) in $y$ -direction?

Velocity along $x$ direction is
$v_{x} = v_{02} + at$
$v_{x} = 3 + (- 1) t$
so when particle reach maximum $x$ coordinate $v_{x} = 0 \Rightarrow t = 3 sec$
similarly $v_{y} = v_{0 y} + a_{y} t$
$v_{y} = 0 + (- \frac{1}{2}) t$
$(v_{y})_{a t t = 3} = - \frac{3}{2} = - 1.5 m s^{- 1}$