Physics, asked by Adhiiiiiiiiiiii, 4 months ago


The acceleration of a particle is given by a(t) = 3t^2 – 6t - 1 m/s2. The particle starts at t = 0 from
position x = 0 with a velocity v = 3 m/s. Find (i) the displacement of the particle, (ii) the distance
covered at the end of 2 s.​

Answers

Answered by maliknuha
1

Answer:

a=3t

2

+2t+2

dv=adt

v=

3

3t

3

+

2

2t

2

+2t

v=t

3

+t

2

+2t

at t=0,u=2

att=2

v=t

3

+t

2

+2t+2

=8+4+2×2+2

=8+4+4+2

=18m/s

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