A particle travels with speed 50 m/s from the point (3 m, -7 m) in a direction 7i^−24j^. Find its position vector after 3 seconds.

Given speed v=50m/s in the direction a→=7i^−24j^v→=va^;a^ is the unit vector in the direction 7i^−24j^a^=(7i^−24j^)(24)2+(7)2=(7i^−24j^)25v→=va^=50⋅(7i^−24j^)25=2(7i^−24j^)m/s Initially the particle is at r→0=(3i^−7j^)m Position of the particle after 3sec,r→=r→0+v→t⇒ r→=(3i^−7j^)+3×2(7i^−24j^)=(45i^−151j^)m