Question #101194
to calculate velocity of a car accelerating from rest in straight line. Equation is v(t) = A (1-e ^ -t/t max speed) the ^ means to the power of. v(t) is the instantaneous velocity of the car (m/s), t is the time in seconds, tmaxspeed is the time to reach the maximum speed in seconds and A is a constant. 1. Create a graph of acceleration vs time? 2. Apply the mathematical model which is a car with a t(0-28m/s) of 2.6s, a t(400m) of 10.46s in order to calculate the value of coefficient, maximum velocity and maximum acceleration?
1
Expert's answer
2020-01-10T08:34:21-0500
v(t)=A(1ettmax_speed)\text v(t)=A(1-e^{-{t \over t_{max\_ speed}}})

a(t)=v(t)=Atmax_speedettmax_speed\text a(t)=\text v'(t)={A \over t_{max\_ speed}}e^{-{t \over t_{max\_speed}}}
v(t)=A(1ettmax_speed)\text v(t)=A(1-e^{-{t \over t_{max\_ speed}}})

s(t)=v(t)dt=A(1ettmax_speed)dt=\text s(t)=\int \text v(t) dt=\int A(1-e^{-{t \over t_{max\_ speed}}})dt=

=A(t+tmax_speedettmax_speed)+C=A(t+t_{max\_speed}\cdot e^{-{t \over t_{max\_ speed}}})+C

s(0)=0=>C=Atmax_speed\text s(0)=0=>C=-A\cdot t_{max\_speed}

s(t)=A(ttmax_speed+tmax_speedettmax_speed)\text s(t)=A(t-t_{max\_speed}+t_{max\_speed}\cdot e^{-{t \over t_{max\_ speed}}})

Given that

v(2.6)=A(1e2.6tmax_speed)=28\text v(2.6)=A(1-e^{-{2.6 \over t_{max\_ speed}}})=28

s(10.46)=\text s(10.46)=A(10.46tmax_speed+tmax_speede10.46tmax_speed)=400A(10.46-t_{max\_speed}+t_{max\_speed}\cdot e^{-{10.46 \over t_{max\_ speed}}})=400


10.46tmax_speed+tmax_speede10.46tmax_speed1e2.6tmax_speed=40028{10.46-t_{max\_speed}+t_{max\_speed}\cdot e^{-{10.46 \over t_{max\_ speed}}} \over 1-e^{-{2.6 \over t_{max\_ speed}}}}={400 \over 28}

tmax_speed5.834 st_{max\_speed}\approx5.834\ s or tmax_speed28.424 st_{max\_speed}\approx28.424 \ s

A77.864 m/sA\approx77.864\ m/s or A320.318 m/sA\approx320.318\ m/s


vmax=A(1etmax_speedtmax_speed)=A(1e1)\text v_{max}=A(1-e^{-{ t_{max\_ speed} \over t_{max\_ speed}}})=A(1-e^{-1})


amax=Atmax_speedetmax_speedtmax_speed=Ae1tmax_speed\text a_{max}={A \over t_{max\_ speed}}e^{-{t_{max\_speed} \over t_{max\_speed}}}={Ae^{-1} \over t_{max\_ speed}}

tmax_speed5.834 s,A77.864 m/st_{max\_speed}\approx5.834\ s,A\approx77.864\ m/s


vmax=77.864(1e1) m/s49.219 m/s\text v_{max}=77.864(1-e^{-1})\ m/s\approx49.219\ m/s

amax=77.864 m/se15.834 s4.910 m/s2\text a_{max}={77.864\ m/s\cdot e^{-1} \over 5.834\ s}\approx4.910\ m/s^2

Or

tmax_speed28.424 s,A320.318 m/st_{max\_speed}\approx28.424\ s,A\approx320.318\ m/s


vmax=320.318(1e1) m/s202.480 m/s\text v_{max}=320.318(1-e^{-1})\ m/s\approx202.480\ m/s

amax=320.318 m/se128.424 s4.146 m/s2\text a_{max}={320.318\ m/s\cdot e^{-1} \over 28.424\ s}\approx4.146\ m/s^2

202.480 m/s729 km/h202.480\ m/s\approx729\ km/h

I think, that the car cannot have such speed.



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