A particle has initial velocity 4i^+4j^ ms-1 and an acceleration -0.4i^ ms-2, at what time will its speed be 5 ms-1?

Since acceleration is in r-direction only, velocity in y-direction will not change. When speed = 5 ms-1    52=Vx2+Vy2=Vx2+42⇒Vx=±3ms−1∴ Vx=ux+axt⇒t=Vx−uxaxor   t1=3−4−0.4=2.5sand t2=-3-4-0.4=17.5s