solution;
given data
initial velocity(u)=0m/s
final velocity(v)=8m/s
time(t)=4sec
according to first equation of motion
v=u+atv=u+atv=u+at
acceleration
a=v−uta=\frac{v-u}{t}a=tv−u
putting the value
a=8−04=2m/s2a=\frac{8-0}{4}=2m/s^2a=48−0=2m/s2
therefore acceleration of runner is 2m/s22m/s^22m/s2 .
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