We use the definition for the acceleration to solve the equation and find the value for k:
a=dtdv=dxdvdtdx=vdxdv=xkxk=vdxdv⟹k∫xdx=∫vdv
We have to use the limits for the integration to calculate k:
k∫x1=7.5mmx2=750mmxdx=∫v1=0m/sv2=600m/svdv⟹k[lnx]x1=7.5mmx2=750mm=21[v2]v1=0m/sv2=600m/s∴k=2ln(7.5mm750mm)(600m/s)2−(0m/s)2 k=39086.50337s2m2
Now we substitute the value for k and x = 375 mm = 0.375 m to find the acceleration of the bullet at that point:
a=xk=0.375m39086.50337s2m2∴ax=375mm=104230.676s2m
Reference:
- Sears, F. W., & Zemansky, M. W. (1973). University physics.
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