A car accelerates with the acceleration a(s) = k · s^n m/s^2, where s is distance in meters, k = 1.8 and n = 0.725. (Unless otherwise stated, assume that the car has a negligible initial speed, which only means that the answer does not become trivial and that the car starts at s = 0.) How far has the car driven when it has reached a speed of 1.4 m/s?
v2=vo2+2asv^2=v_o^2+2asv2=vo2+2as
v2=2asv^2=2asv2=2as
1.42=2∗k∗sn∗s1.4^2=2*k*s^n*s1.42=2∗k∗sn∗s
1.42=2⋅ 1.8ss0.7251.4^2=2\cdot \:1.8ss^{0.725}1.42=2⋅1.8ss0.725
1.42=3.6s0.725s1.4^2=3.6s^{0.725}s1.42=3.6s0.725s
Use the following exponent property:an=(a1m)(n⋅ m)\mathrm{Use\:the\:following\:exponent\:property}:\quad a^n=\left(a^{\frac{1}{m}}\right)^{\left(n\cdot \:m\right)}Usethefollowingexponentproperty:an=(am1)(n⋅m)
1.42=3.6u29u401.4^2=3.6u^{29}u^{40}1.42=3.6u29u40
s=(1.963.6)4069s=\left(\frac{1.96}{3.6}\right)^{\frac{40}{69}}s=(3.61.96)6940
1.42=3.6(s140)29(s140)401.4^2=3.6\left(s^{\frac{1}{40}}\right)^{29}\left(s^{\frac{1}{40}}\right)^{40}1.42=3.6(s401)29(s401)40
s=(1.963.6)4069=s=0.70295ms=\left(\frac{1.96}{3.6}\right)^{\frac{40}{69}}=s=0.70295 ms=(3.61.96)6940=s=0.70295m
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