Question

plzz solve no. 15 in the following image

Answer 1

i) Let max velocity be =v
   u=0,a=2 m/s2,t=10 s
   v=u+at
     =0+2*10
     =20 m/s (ANSWER)
ii) Let the retardation be=a
    u=20 m/s,v=0,t=50s
    v=u-at
=>0=20-50a
=>50a=20
=>a=2/5
=>a=0.4 m/s2 (ANSWER)
iii) Case:1
     u=0,a=2 m/s2, t=10s
     S=ut+1/2*a*t*t
       =0*10+1/2*2*10*10
       =0+100
       =100m
Case:2
u=20m/s,t=200s,a=0
S=ut+1/2*a*t*t
  =20*200+1/2*0*200*200
  =4000m
Case:3
u=20m/s,t=50s,a=-0.4m/s2
S=ut+1/2*a*t*t
  =20*50+1/2*4/10*50*50
  =500m
Total distance=100+4000+500
=4600m (ANSWER)
iv)Avg. velocity=Total Distance/Total Time
=4600/260
=17.69 m/s (ANSWER)

Answer 2

i) Let velocity be =v
u=0,a=2 m/s2,t=10 s
v=u+at
=0+2*10
=20 m/s (ANSWER)
ii) Let the retardation be=a
u=20 m/s,v=0,t=50s
v=u-at
=>0=20-50a
=>50a=20
=>a=2/5
=>a=0.4 m/s2 (ANSWER)
iii) Case:1
u=0,a=2 m/s2, t=10s
S=ut+1/2*a*t*t
=0*10+1/2*2*10*10
=0+100
=100m
Case:2
u=20m/s,t=200s,a=0
S=ut+1/2*a*t*t
=20*200+1/2*0*200*200
=4000m
Case:3
u=20m/s,t=50s,a=-0.4m/s2
S=ut+1/2*a*t*t
=20*50+1/2*4/10*50*50
=500m
Total distance=100+4000+500
=4600m (ANSWER)
iv)Avg. velocity=Total Distance/Total Time
=4600/260
=17.69 m/s