Answer:
the acceleration
![a^(\to) = (0.0159 \ \ m/s^2 )i](https://img.qammunity.org/2021/formulas/physics/college/y80n2yhhi1orkaoqfinjaw9o9txakj3dvv.png)
Step-by-step explanation:
Given that:
the initial speed v₁ = 0 m/s i.e starting from rest ; since the car accelerates at a distance Δx = 6 miles in order to teach that final speed v₂ of 63.15 km/h.
So; the acceleration for the first 6 miles can be calculated by using the formula:
v₂² = v₁² + 2a (Δx)
Making acceleration a the subject of the formula in the above expression ; we have:
v₂² - v₁² = 2a (Δx)
![a = (v_2^2 - v_1^2 )/(2 \Delta x)](https://img.qammunity.org/2021/formulas/physics/college/bwnmw4o1ppl949sh5srjfhaowqv83ct2sg.png)
![a = ((63.15 \ km/s)^2 - (0 \ m/s)^2 )/(2 (6 \ miles))](https://img.qammunity.org/2021/formulas/physics/college/plh4m10eavirfpt7lkellaafyh126efxv1.png)
![a = ((17.54 \ m/s)^2 - (0 \ m/s)^2 )/(2 (9.65*10^3 \ m))](https://img.qammunity.org/2021/formulas/physics/college/2j1686idb762k1spy7zqv3qz34llrtvn13.png)
![a =0.0159 \ m/s^2](https://img.qammunity.org/2021/formulas/physics/college/dyqls7yh6d3geciv8kpptefbc73gpz6w5g.png)
Thus;
Assume the car moves in the +x direction;
the acceleration
![a^(\to) = (0.0159 \ \ m/s^2 )i](https://img.qammunity.org/2021/formulas/physics/college/y80n2yhhi1orkaoqfinjaw9o9txakj3dvv.png)