Question

Velocity
and acceleration of a particle at some instant of time
are v = (3i +4j) m/s and a=-(6i +8j)m/s^2
respectivly.at the same instant particle is at origin.maximum x coordinate of particle will be​

Answer 1

Answer:

3/4

Explanation:

Velocity

and acceleration of a particle at some instant of time

are v = (3i +4j) m/s and a=-(6i +8j)m/s^2

respectively.at the same instant particle is at origin.maximum x coordinate of particle will be​

S = ut + (1/2)at²

S = (3i + 4j)t + (1/2)(-(6i + 8j)t²

S = (3i + 4j)t - (3i + 4j)t²

S = (3i + 4j)(t - t²)

for Max x coordinate

f(t) = t - t²  should be max

Differentiating with  t

f'(t) = 1 - 2t = 0

=> t = 1/2

Differentiating again

f''(t) = - 2

=> t = 1/2 will Give Max value

S = (3i + 4j)(t - t²) at t = 1/2

S =  (3i + 4j)(1/2 - 1/4)

S =  (3i + 4j)/4

Max x coordinate would be 3/4