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Grade 12Mechanics

Velocity of particle at time t=0 is 2m/s. A constant acceleration of 2m/s^2 acts on the particle for 1 sec at angle of 60° with it`s initial velocity. find the magnitude of velocity at the end of 1 second.

Profile image of Pata venkateswarlu
8 Years agoGrade 12
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3 Answers

Profile image of Sahil kumar
8 Years ago
Acceleration in x direction is 2cos60 = 1m/s2, Acceleration in y direction is 2sin60 = 31/2m/s2. After 1 sec VX = 2 + 1(1) = 3m/s and Vy= 0 + 31/2(1). and then find resultant velocity which gives you magnitude at the end of 1 sec.
 
Profile image of Shivam tiwari
8 Years ago
Direction along the initial velocity: Vo = 2m/s, a = 2m/s^2 * cos(60) = 1m/s^2, t = 2 second Use Vf = Vo + a * t = 2m/s + 1m/s^2 * 2s = 4m/s. Use S = Vo*t + (1/2) * a * t^2 = 2m/s * 2s + (1/2) * 1m/s^2 * (2s)^2 = 4m + 2m = 6m Direction perpendicular to the initial velocity: Vo =0m/s, a = 2m/s^2 * sin(60) = 1.732m/s^2, t = 2 second Again, Use Vf = Vo + a * t = 0 + 1.732m/s^2 * 2s = 3.46m/s. Use S = Vo*t + (1/2) * a * t^2 = 0 + (1/2) *1.732m/s^2 * (2s)^2 = 3.46m Now, combine velocities in both direction (Vectorially) , and combine diplacement in both direction (Vectorially). This left for you to be done.
Profile image of yamun sharma
6 Years ago
u= 2î
a= (2cos60°)î +(2sin60°)j 
=(î + √3j)
t= 1s.
 
v= u + at = (2î) + (î + √3j) (1)
= 4î + √3j
 
|v| = √(4)2 + (√3)2
=√19
= 4.35 m/s.
 
|v| = √()2
 

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