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Question Number 14219 by tawa tawa last updated on 29/May/17
Answered by ajfour last updated on 29/May/17
s¯=4t3i^−2t2j^+3tk^..(i)(a)v¯=ds¯dt=12t2i^−4tj^+3k^...(ii)(b)speedatt=1s=∣v¯∣t=1s=144+16+9=13(c)a¯=dv¯dt=24ti^−4j^.....(iii)att=3s,a¯=72i^−4j^(d)att=0,fromeqn(ii):v¯0=3k^.
Commented by tawa tawa last updated on 29/May/17
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