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Solution 8.8b

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We must find the position of the particle after 30 s.

Using r=ut+21at 2+r0 with  a=i+2j ms-2 which was calculated in part a),  u=4i+6j ms-1 and  r0=0 gives,

r=(4i+6j)30+21(i+2j)30 2

r=120i+180j+450i+900j=570i+1080j

As the starting point is at the origin, the distance is the magnitude of this position vector.

5702+1082=324900+1166400=1491300=1221 m