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.
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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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