Question

A particle starts from rest with uniform acceleration. Its displacement after t seconds is given by

x = 3t2 + 6t + 9 (in metres)

Calculate the magnitude of its (a) initial velocity, (b) velocity when t =2s, (c) initial acceleration and (d) displacement

when t = 4s.​

Answer 1

Answer:

a) v=6t+6

b) v(2)=18m/s

c) a=6m/s2

d) x(4)=78m

Explanation:

velocity=v=dx/dt

a) v=6t+6

b) v(2)= 6(2)+6

=12+6

=18m/s

c) a=dv/dt

a=6m/s2

d) x(4)=3t2 + 6t + 9 (in metres)

x(4)=3(4)2+6(4)+6

x(4)=3(16)+24+6

x(4)=48+24+6

x(4)=78m

Answer 2

Answer:

Given equation of path of particle is x=7t

2

+5t+2

Now,

v=

dt

dx

=14t+5.

And initial velocity is the velocity of particle at time t=0. Hence ;

v

i

=14(0)+5=5ms

−1

So the magnitude of Initial velocity is 5ms

−1

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