A point moves in a straight line so that its distance from the start in time t is equal to s=1/4t^4-4t^3+16t^2 find:

a) At what times was the point at its starting position?
b) At what times its velocity equal to zero?

Dear Student,

Please find below the solution to the asked query:

The distance travelled by the particle varies with time as,

S=14t4-4t3+16t2
So, if the particle should be at the starting point. Then,

S=014t4-4t3+16t2=0t2-16t+64=0t2-8t-8t+64=0t=8 sec

The velocity of the particle is,

V=dSdtV=ddt14t4-4t3+16t2V=t3-12t2+32tIf V=0t3-12t2+32t=0t2-12t+32=0t2-8t-4t+32=0tt-8-4t-8=0t-4t-8=0t = 4 s & t = 8 s

 

Hope this information will clear your doubts about the topic.

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