Question

A Particle Starts Moving With Initial Velocity 5M/S. It Is Accelerating With A Constant Acceleration 2M/S At An Angle 120 Degree With The Direction Of Initial Veleocity. Find The Time After Which Its Speed Would Be √3 Times That Of The Initial Speed​

Answer 1

Answer:

5 Sec

Explanation:

A Particle Starts Moving With Initial Velocity 5M/S. It Is Accelerating With A Constant Acceleration 2M/S At An Angle 120 Degree With The Direction Of Initial Veleocity. Find The Time After Which Its Speed Would Be √3 Times That Of The Initial Speed​

Let say Initial Velcoity

= 5 i + 0j

acceleration = 2m/s² at an angle of 120°

a = 2Cos120i + 2Sin120j

=> a = - i + √3 j

Let say after t time its Speed become √3 Times That Of The Initial Speed​

Then after t time using V = U + at

Velocity = (5 i + 0j) + (- i + √3 j)t

= (5 - t)i + t√3 j

Magnitude = √(5-t)² + (t√3)²  = √(25 + t² - 10t + 3t²)

= √4t² - 10t + 25

as it become √3 times of initial speed

√4t² - 10t + 25 = 5√3

Squaring both side

4t² - 10t + 25 = 75

=> 4t² - 10t - 50 = 0

=> 2t² - 5t - 25 = 0

=> 2t² - 10t +5t - 25 = 0

=> 2t(t - 5) + 5(t - 5) = 0

=> (2t + 5)(t - 5) = 0

=> t = 5   ( as time can not be -ve)

After 5 Secs Speed Would Be √3 Times That Of The Initial Speed​