The coordinates of a particle moving in x-y plane at any instant of time t are x = 4t2 ; y = 3t2. The speed of the particle at that instant is

Question

The coordinates of a particle moving in x-y plane at any instant of time t are x = 4t2 ; y = 3t2. The speed of the particle at that instant is (A) 10 t (B) 5 t (C) 3 t (D) 2 t

Tardigrade Admin · Accepted Answer

x=4 t2 Differentiate with respect to t, we get vx=(d x/d t)=2 × 4 t=8 t Similarly, vY=(d y/d t)=2 × 3 t=6 t The speed of the particle V =√VX2+VY2 =√(8 t)2+( overline6 t)2 =√61 t2+36 t2 =√100 t2=1 0 t

Q. The coordinates of a particle moving in $x - y$ plane at any instant of time $t$ are $x = 4 t^{2}; y = 3 t^{2}$ . The speed of the particle at that instant is

10 t

5 t

3 t

2 t

$13 t$

$x = 4 t^{2}$
Differentiate with respect to $t$ , we get
$v_{x} = \frac{d x}{d t} = 2 \times 4 t = 8 t$
Similarly, $v_{Y} = \frac{d y}{d t} = 2 \times 3 t = 6 t$
The speed of the particle
$V = V_{X}^{2} + V_{Y}^{2}$
$= (8 t)^{2} + (\overline{6} t)^{2}$
$= 61 t^{2} + 36 t^{2}$
$= 100 t^{2} = 10 t$

Q. The coordinates of a particle moving in x−y plane at any instant of time t are x=4t2;y=3t2. The speed of the particle at that instant is

10 t

5 t

3 t

2 t

13​t

x=4t2 Differentiate with respect to t, we get vx​=dtdx​=2×4t=8t Similarly, vY​=dtdy​=2×3t=6t The speed of the particle V=VX2​+VY2​​ =(8t)2+(6t)2​ =61t2+36t2​ =100t2​=10t

Q. The coordinates of a particle moving in $x - y$ plane at any instant of time $t$ are $x = 4 t^{2}; y = 3 t^{2}$ . The speed of the particle at that instant is

$13 t$

$x = 4 t^{2}$
Differentiate with respect to $t$ , we get
$v_{x} = \frac{d x}{d t} = 2 \times 4 t = 8 t$
Similarly, $v_{Y} = \frac{d y}{d t} = 2 \times 3 t = 6 t$
The speed of the particle
$V = V_{X}^{2} + V_{Y}^{2}$
$= (8 t)^{2} + (\overline{6} t)^{2}$
$= 61 t^{2} + 36 t^{2}$
$= 100 t^{2} = 10 t$