The given situation of entire motion of motor bike is shown in the figure. If time taken by the motorbike to reach from point A to B is tAB, then for A to B, vA=0,vB=10ms−1,a=0.5ms−2 ∴ From equation,v=u+at vB=vA+atAB⇒10=0+0.5tAB ⇒ tAB=‌
10
0.5
=20s Since, motorbike moves from point B to C with constant velocity. Hence, time taken to travel distance BC is given as tBC=‌
BC
vB
=‌
10km
10ms−1
=‌
10000m
10ms−1
=1000s For motion of motorbike from point C to D, time =tCD,vB=10ms−1,vf=0,a′=0.2ms−2 From equation, v=u−at, vf=vB−a′tCD 0=10−0.2×tCD⇒tCD=50s ∴‌ Total time ‌=‌‌tAB+tBC+tCD =‌‌20s+1000s+50s=1070s