Question

Q. Comment on acceleration (const an or
variable ) of motion if position variation is
given.
a) 2 = 3t -2
b) x = - 4 t^3 + 8t² - 6
c) x = 5t² – 3​

Answer 1

Answer:

option B is correct.because a is position and c is velocity. so..........

Answer 2

Acceleration is obtained by double diffrenciation of distance

1)x=3t-3

V=3

A=0

2) x=-4t^3 +8t²-6

V=-12t²+16t-0

A=-24t+16

Acc is variable it changes with time

3)x=5t²-t

V=10t-0

A=10

Acc is const

I hope this helps ( ╹▽╹ )