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A particle moves along the X-axis as x = u(t - 2s) + a(t - 2s)^2. What is true?..................a) the initial velocity is u. b) the acc. is a. c) the acc. is 2a. d) at t = 2s, particle is at origin. Will the asnwer be a,c and d?

A particle moves along the X-axis as x = u(t - 2s) + a(t - 2s)^2. What is true?..................a) the initial velocity is u. b) the acc. is a. c) the acc. is 2a. d) at t = 2s, particle is at origin. Will the asnwer be a,c and d?

Grade:11

4 Answers

Mukul Tuli
18 Points
12 years ago

answer is a ,b an d

AskIITianExpert Srijan.iitd
8 Points
12 years ago

yes the answer will be a c and d assuming constant acceleration.

since x is the coordinate point and particle moves only along the x axis, hence x is the displacement of particle.

now comparing the expression with the 2nd equation of motion we can easily conclude u is the initial velocity because its the coefficient of first degree term(expressinon being a function of time shifted by 2 units).

and acceleration being 2a.

dharani bandaru
18 Points
11 years ago

I think the answer is c and d.when the given relation is differentiated once we get the velocity and when u put t=0 velocity will not be u.so "a" is wrong.again differentiate it .we will get the acceleration .put t=0 .accel. is 2a      so "b"is wrong.and "c" is correct. coming to "d" put t=0 in given relation x=0 is the answer . dat means it is at the origin.so "d" is true.

Amarjit
105 Points
4 years ago
On differentiating x with respect to t we get the equation; dx/dt= u+2a(t-2s).dx/dt is velocity and here it is in terms of t.Initial velocity is the velocity at t=0. So, on putting t=0, we get dx/dt(t=0)= u-4as.Hence option a is not correntCorrect answer c,d.

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