Question

The instantaneous coordinates of a particle are x=6t and y=4t^2 metre .what will be the velocity of the particle at t=1sec ?

Answer 1

Given:

The coordinates of a particle are x=6t and y=4t² metre.

To find:

Velocity at t = 1 sec?

Calculation:

$1) \: x = 6t$

$\implies \: v_{x} = \dfrac{dx}{dt}$

$\implies \: v_{x} = \dfrac{d(6t)}{dt}$

$\implies \: v_{x} = 6 \: m {s}^{ - 1}$

Now, in Y axis:

$2) \: y= 4 {t}^{2}$

$\implies \: v_{y} = \dfrac{dy}{dt}$

$\implies \: v_{y} = \dfrac{d(4 {t}^{2} )}{dt}$

$\implies \: v_{y} = 8t$

$\implies \: v_{y} = 8 \times 1 = 8 \: m {s}^{ - 1}$

So, net velocity is :

$v = \sqrt{ {6}^{2} + {8}^{2} } = \sqrt{100} = 10 \: m {s}^{ - 1}$

So, net velocity is 10 m/s.

Answer 2

Explanation:

1)x=6t

\implies \: v_{x} = \dfrac{dx}{dt} ⟹v

x

=

dt

dx

\implies \: v_{x} = \dfrac{d(6t)}{dt} ⟹v

x

=

dt

d(6t)

\implies \: v_{x} = 6 \: m {s}^{ - 1} ⟹v

x

=6ms

−1

Now, in Y axis:

2) \: y= 4 {t}^{2} 2)y=4t

2

\implies \: v_{y} = \dfrac{dy}{dt} ⟹v

y

=

dt

dy

\implies \: v_{y} = \dfrac{d(4 {t}^{2} )}{dt} ⟹v

y

=

dt

d(4t

2

)

\implies \: v_{y} = 8t⟹v

y

=8t

\implies \: v_{y} = 8 \times 1 = 8 \: m {s}^{ - 1} ⟹v

y

=8×1=8ms

−1

So, net velocity is :

v = \sqrt{ {6}^{2} + {8}^{2} } = \sqrt{100} = 10 \: m {s}^{ - 1} v=

6

2

+8

2

=

100

=10ms

−1

So, net velocity is 10 m/s.