Question

16.The displacement x of a particle varies with time as x = 4t2 – 15t + 25. (i) Find the position, velocity and acceleration of the particle at t = 0. (ii) Find average velocity during 2 s to 4 s.

Answer 1

when t=0

x=25m

by using differeciation nxn-1

velocity=dx/dt= 2*4*t-1 *15



v=dx/dt =8t-15

t=o

v=-15m/s

a=dv/dt

v=8t-15

=8-0

=8m/s2

Answer 2

To find position of particle put t=0 in given equation we get x=25

to find velocity differentiate with respect to t we get dx/dt=8t-15 now put t=0 we get velocity= - 15

now to find acceration differentiate equation of velocity w.r.t t we get d^2 x/dt^2=8.

As acceleration remains constant it is uniform acceleration

now,consider equation of velocity dx/dt=8t-15

put v=0 we get 8t=15 t=15/8

at t=15/8 velocity will be zero

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