Accelerated motion - mechanics

The delivery truck with a total weight of 3.6 t accelerates from 76km/h to 130km/h in the 0.286 km long way. How much was the force needed to achieve this acceleration?

Result

F =  5402.098 N

Solution:

m=3.6 t=3.6⋅ 1000 kg=3600 kg s=0.286 km=0.286⋅ 1000 m=286 m v1=76 km/h=76/3.6 m/s=21.11111 m/s v2=130 km/h=130/3.6 m/s=36.11111 m/s  s=v1⋅ t+12at2 a=v2−v1t  s=v1⋅ t+12v2−v1tt2 s=v1⋅ t+12(v2−v1)t  t=sv1+(v2−v1)/2=28621.1111+(36.1111−21.1111)/2≐5148515≐9.9961 s  a=v2−v1t=36.1111−21.11119.9961≐1.5006 m/s2  s1=v1⋅ t+12⋅ a⋅ t2=21.1111⋅ 9.9961+12⋅ 1.5006⋅ 9.99612=286 m s1=s  F=m⋅ a=3600⋅ 1.5006≐5402.0979≐5402.098 Nm=3.6 \ t=3.6 \cdot \ 1000 \ kg=3600 \ kg \ \\ s=0.286 \ km=0.286 \cdot \ 1000 \ m=286 \ m \ \\ v_{1}=76 \ km/h=76 / 3.6 \ m/s=21.11111 \ m/s \ \\ v_{2}=130 \ km/h=130 / 3.6 \ m/s=36.11111 \ m/s \ \\ \ \\ s=v_{1} \cdot \ t + \dfrac{ 1 }{ 2 } a t^2 \ \\ a=\dfrac{ v_{2}-v_{1} }{ t } \ \\ \ \\ s=v_{1} \cdot \ t + \dfrac{ 1 }{ 2 } \dfrac{ v_{2}-v_{1} }{ t } t^2 \ \\ s=v_{1} \cdot \ t + \dfrac{ 1 }{ 2 } (v_{2}-v_{1}) t \ \\ \ \\ t=\dfrac{ s }{ v_{1}+(v_{2}-v_{1})/2 }=\dfrac{ 286 }{ 21.1111+(36.1111-21.1111)/2 } \doteq \dfrac{ 5148 }{ 515 } \doteq 9.9961 \ \text{s} \ \\ \ \\ a=\dfrac{ v_{2}-v_{1} }{ t }=\dfrac{ 36.1111-21.1111 }{ 9.9961 } \doteq 1.5006 \ \text{m/s}^2 \ \\ \ \\ s_{1}=v_{1} \cdot \ t + \dfrac{ 1 }{ 2 } \cdot \ a \cdot \ t^2=21.1111 \cdot \ 9.9961 + \dfrac{ 1 }{ 2 } \cdot \ 1.5006 \cdot \ 9.9961^2=286 \ \text{m} \ \\ s_{1}=s \ \\ \ \\ F=m \cdot \ a=3600 \cdot \ 1.5006 \doteq 5402.0979 \doteq 5402.098 \ \text{N}

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