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# Answer to Question #71811 in Mechanics | Relativity for PULKIT

Question #71811
The position of a particle moving in a straight line is given by x = 3 + 4t + 3t2, where x is in
metre and time is in second. Find the values of the following physical quantities for the particle
at t = 2s.
The particle motion law is given by
x = 3 + 4t + 3t^2
At the moment of time t=2 s
a) Position
x=3+4×2+3×2^2=23 m
b) Velocity
v(t)=x' (t)=4+6t
v(2)=4+6×2=16 m/s
c) Acceleration
a(t)=x'' (t)=6 m/s^2

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Assignment Expert
08.01.18, 22:28

Dear Asha Priyadharshini, you are right.
d = |x(0) - x(2)| = 20 m

08.01.18, 18:14

It's wrong!
Displacement= 23-3= 20m

Assignment Expert
18.12.17, 15:14

Just a difference between starting position x(0) = 3 and final postion x(2) = 23:
d = |x(0) - x(2)| = 21 m.

pulkit narula
17.12.17, 08:03

displacement ?