As per the question,
Initial speed (u)=50km/hr=50×10003600m/sec=1259m/s(u)=50km/hr = 50\times\frac{1000}{3600}m/sec=\frac{125}{9}m/s(u)=50km/hr=50×36001000m/sec=9125m/s
Final speed (v)=0(v)=0(v)=0
Covered distance (d)=60m(d)=60m(d)=60m
Let the acceleration is =a=a=a
Now, v2=u2−2adv^2=u^2-2adv2=u2−2ad
Now, substituting the values,
⇒a=u22d=12522×92×60m/s2\Rightarrow a=\frac{u^2}{2d}=\frac{125^2}{2\times 9^2\times 60}m/s^2⇒a=2du2=2×92×601252m/s2
⇒a=1.60m/s2\Rightarrow a=1.60 m/s^2⇒a=1.60m/s2
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