Answer to Question #135412 in Mechanics | Relativity for Mac Roy

Question #135412
A truck covers 40.0 m in 8.50 s while smoothly slowing down to a final speed of 2.80 m/s. (a) Find its original speed. (b) Find its acceleration.
1
Expert's answer
2020-10-07T10:17:55-0400

B) Since V1=V0a×tV_1=V_0-a\times t , where V1V_1 is the final velocity, V0V_0 is the initial velocity, then V0=V1+a×tV_0=V_1+a\times t ,

and since S=V12V022×aS =\frac {V_1^2-V_0^2}{2\times a} ,

S=V12V12+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}

a=2×S2×V1×tt2=2×402×2.8×8.58.520.45a=\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}\approx0.45 m/s2^2

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=6.625 m/s


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