Question

1.
The velocity of a particle at t = 0 is
u = (4i+4j) ms^-1 and its acceleration is
-(10 j)ms^-2. At t = 0, the particle is at
(1, 0) m. The x - coordinate of the point,
where its y-coordinate is again zero is​

Answer 1

Answer:

The x coordinate of the point where its y coordinate is again zero is 5.

Explanation:

velocity of the particle

$\vec u=4\hat i+4\hat j$

i.e. velocity of the particle in y direction is 4 m/s

Acceleration of the particle

$\vec a=-10\hat j$

i.e. acceleration in the y-direction is -10 m/s²

At t = 0, the y-displacement is zero

Let after time t its displacement is again zero then

Using the second equation of motion in the y-direction

$\boxed{h=ut+\frac{1}{2}at^2}$

$0=4\times t+\frac{1}{2}\times (-10)\times t^2$

$\implies 0=4t-5t^2$

$\implies t(4-5t)=0$

$\implies t=0, \frac{5}{4}$

$\implies t=\frac{5}{4}$ sec

The velocity of the particle in y-direction is 4 m/s

Since there is no acceleration in x-direction

Therefore, the displacement in the x-direction

$=\text{speed}\times \text{time}$

$=4\times\frac{5}{4}}$

$=5$ m

Thus the x coordinate of the point where its y coordinate is again zero is 5.

Hope this helps.

Answer 2

Hint: We only have to deal with the y components of the given vector quantities firstly.

Answer:

x = 4.2

Explanation:

u = 4î + 4ĵ m/s

a = - 10ĵ m/s²

xₒ = (1, 0)

Here,

uᵧ = 4ĵ

aᵧ = - 10ĵ

xₒᵧ = 0ĵ

We know,

∆x = ut + 1/2at²

=> 0 = uᵧt + 1/2aᵧt²

(∆x = 0 because initial y position is 0 and thus displacement is zero when the y coordinate will become 0 again.)

=> 0 = (4)t + 1/2(-10)t²

=> 4t - 5t² = 0

=> t(4 - 5t) = 0

Either, t = 0 (rejected because we already know that y comp of displacement is 0 when t = 0).

or, 5t = 4

=> t = 4/5 s.

Therefore, at t = 0.8 s, the y component is 0.

x component at t = .8:

xₓ = xₒₓ + uₓt + 1/2aₓt²

=> x coordinate = 1 + 4(4/5) = 4.2.