Question

the position of a particle at time x=t^2-4t+6 find its acceleration at 0-3 second and its instantaneous velocity at time 6sec​

Answer 1

Answer:

Given,

x=t2−4t+6

So, the change in velocity is:

dtdx=2t−4

Since velocity is changing,

At t=0,x1=6

At t=2,x2=2

So the magnitude of the distance traveled is:

6−2=4m

At t=3,x3=3

So distance traveled from t=2s to t=3s

x3−x2=3−2=1m

Thus the total distance is:

4+1=5m

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