nitial velocity= 80 km/h = 22.2 m/s v= final velocity = 60 km/h = 16.7 m/s t= time = 5 s a= acceleration = ? a=v-u/t a= 16.7-22.2/5 a= -5.5/5 Therefore, a=-1.1 2.The TRAIN started from a railway station, so initial velocity of train, u = 0 Final velocity of train, v = 40 kmph  Final velocity, v = 40 × 5/18 = 11.11 m/s Assuming uniform acceleration of ‘a’ m/second square Time taken by train to get it's final velocity, t = 10 minutes = 600 seconds Using first equation of motion, V = u + at 11.11 = 0 + a×600 a = 11.11/600 = 0.0185 m/second square. So, required acceleration , a = 0.0185 3.Average speed=total DISTANCE/total time d=distance travelled at 30m/s and distance travelled at 40m/s … same distance time=distance/speed time at 30m/s =d/30 time at 40 m/s=d/40 total time=d/30 + d/40= 7d/120 average speed= (d+d)/(7d/120)= 240/7=34.2857 m/s It is not the average of the speeds because he spends more time at 30 m/s then at 40 m/s4.Distance covered in half an hour at 30km/hr = s×t = 30×1/2=15km.Distance COVETED in 1hr at 25km/hr= s×t= 25×1 =25km.Distance covered in 2 hrs at 40km/hr=s×t= 40×2=80kmTherefore, total distance covered= 15+25+80=120kmTotal time taken= 3HRS and 30 mins=7/2 hrs.Therefore, average speed=total distance covered/total time taken = 120×2/7 = 34.28km/hr5.