B) Since V1=V0−a×tV_1=V_0-a\times tV1=V0−a×t , where V1V_1V1 is the final velocity, V0V_0V0 is the initial velocity, then V0=V1+a×tV_0=V_1+a\times tV0=V1+a×t ,
and since S=V12−V022×aS =\frac {V_1^2-V_0^2}{2\times a}S=2×aV12−V02 ,
S=V12−V12+2×V1×a×t+a2×t22×aS=\frac {V_1^2-V_1^2+2\times V_1\times a \times t + a^2 \times t^2}{2\times a}S=2×aV12−V12+2×V1×a×t+a2×t2
a=2×S−2×V1×tt2=2×40−2×2.8×8.58.52≈0.45a=\frac{2\times S- 2\times V_1 \times t }{t^2}=\frac{2\times 40- 2\times 2.8 \times 8.5}{8.5^2}\approx0.45a=t22×S−2×V1×t=8.522×40−2×2.8×8.5≈0.45 m/s2^22
а) V0=V1+a×t=2.8+0.45×8.5=6.625V_0=V_1+a\times t=2.8+0.45\times 8.5=6.625V0=V1+a×t=2.8+0.45×8.5=6.625 m/s
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