Question

The acceleration of a particle moving in the xy plane is given by a = (3i + 4j) m/s² at t = 0 the position vector of the particle r0 = (20i + 40j) m and the velocity is u= (5i + 2j) m/s after 4 s determine its 1 displacement 2 position vector 3 the magnitude of velocity 4 the angle between its direction of travel and the + X axis​

Answer 1

Answer:

hey mate

Step-by-step explanation:

here is your answer

Given

r0 = ( 20i + 40ij) m

a = (3i + 4j) m/s²

u = (5i + 2j) m/s t = 4s

solution

1. The displacement of the particle is 4 s

s = ut + 1/2 at²

= (5i + 2j) (4) + 1/2 (5i + 4j) (4)²

= (20i + 8j) + (24i + 32j) = (44i + 40j) m

= 4(11i + 10j)

2. s = r - r0

the position vector after 4 s

r = r0 + s (20i + 40j) + (44i + 40j)

= (64i + 8j) = 16 (4i + 5j)m

= (64i + 8j)

= 16(4i + 5j)m

3. the instantaneous velocity after 4 s

v = u + at = (5j + 2j) + (3i + 4j) (4)

= (5+12)i + (5 + 16)j (17i + 18j) m/s

The magnitude of the instantaneous velocity

$v \: = \: |v | \sqrt{ {v}^{2}x \: + {v}^{2}y } = \sqrt{ {(17)}^{2} + {(18)}^{2} }$

$\sqrt{289 + 324} = \sqrt{ {(17)}^{2} + {(18)}^{2} }$

$\sqrt{289 + 324}$

$= \sqrt{613}$

= 24.76 m/s

and v makes an angle theta with the + axis given by

$\tan(theta) = \frac{vy}{vx} = \frac{18}{17} = 1.0588$

theta = tan-1 1.0588

= 46°38

$thanku$

Answer 2

refer to the attachment........

be brainly ✌️