Question

A particle is moving in x-y plane. Its initial velocity and acceleration are u=(4 hati+8 hatj) m//s and a=(2 hati-4 hatj) m//s^2. Find (a) the time when the particle will cross the x-axis. (b) x-coordinate of particle at this instant. (c) velocity of the particle at this instant. Initial coordinates of particle are (4m,10m).

Answer 1

answer : (a) 5sec (b) 49m (c) 14i - 12 j

given, initial velocity of particle, u = (4i + 8j) m/s

acceleration of particle,a = (2i - 4j) m/s²

initial co-ordinates of the particle = (4, 10)

(a) time when particle crosses the x-axis.

i.e., y = 0

using formula, $y=y_0+u_yt+\frac{1}{2}a_yt^2$

here, y = 0, $u_y=8,a_y=-4, y_0=10$

so, 0 = 10 + 8t - 1/2 × 4 × t²

⇒0 = 10 + 8t - 2t²

⇒2t² - 8t - 10 = 0

⇒t² - 4t - 5 = 0

⇒t² - 5t + t - 5 = 0

⇒t = 5 sec

(b) x - co-ordinates of particle at this instant.

using formula, $x=x_0+u_xt+\frac{1}{2}a_xt^2$

here, $x_0=4,u_x=4,a_x=2,t=5$

so, x = 4 + 4(5)+ 1/2 × 2 × (5)²

= 4 + 20 + 25

= 49 m

(c) velocity of particle at this instant.

$\vec v=\vec u+\vec a t$

= (4i + 8j) + (2i - 4j) × 5

= 4i + 8j + 10i - 20j

= 14i - 12j

Answer 2

answer : (a) 5sec (b) 49m (c) 14i - 12 j

given, initial velocity of particle, u = (4i + 8j) m/s

acceleration of particle,a = (2i - 4j) m/s²

initial co-ordinates of the particle = (4, 10)

(a) time when particle crosses the x-axis.

i.e., y = 0

using formula, y=y_0+u_yt+\frac{1}{2}a_yt^2y=y

0

+u

y

t+

2

1

a

y

t

2

here, y = 0, u_y=8,a_y=-4, y_0=10u

y

=8,a

y

=−4,y

0

=10

so, 0 = 10 + 8t - 1/2 × 4 × t²

⇒0 = 10 + 8t - 2t²

⇒2t² - 8t - 10 = 0

⇒t² - 4t - 5 = 0

⇒t² - 5t + t - 5 = 0

⇒t = 5 sec

(b) x - co-ordinates of particle at this instant.

using formula, x=x_0+u_xt+\frac{1}{2}a_xt^2x=x

0

+u

x

t+

2

1

a

x

t

2

here, x_0=4,u_x=4,a_x=2,t=5x

0

=4,u

x

=4,a

x

=2,t=5

so, x = 4 + 4(5)+ 1/2 × 2 × (5)²

= 4 + 20 + 25

= 49 m

(c) velocity of particle at this instant.

\vec v=\vec u+\vec a t

v

=

u

+

a

t

= (4i + 8j) + (2i - 4j) × 5

= 4i + 8j + 10i - 20j

= 14i - 12j