91Ó°ÊÓ

Using the given velocity function \(v=2t^2\), we know that the car's tangential velocity is given by \(v_t = 2t^2\). Since the car is moving in a circular path, its radial velocity \(v_r\) will be zero.

To find the tangential acceleration, we need to find the first derivative of the tangential velocity function: \(a_t = \frac{dv_t}{dt}\). Differentiating the function \(v_t = 2t^2\), we get: \[a_t = \frac{d(2t^2)}{dt} = 4t\]

The centripetal acceleration is given by the formula \(a_c = \frac{v_t^2}{r}\), where \(r\) is the radius of the circular path. We already have \(v_t = 2t^2\) and \(r = 100~\text{m}\). Plugging in the values, we get: \[a_c = \frac{(2t^2)^2}{100}\]

The total acceleration of the car is the vector sum of the tangential acceleration and centripetal acceleration. Their magnitudes can be combined using the Pythagorean theorem: \[a = \sqrt{(a_t)^2 + (a_c)^2}\] Since we want to find the acceleration at \(t = 5~\text{s}\), plug in the value of \(t\) in the expressions for \(a_t\) and \(a_c\): \[a_t = 4(5) = 20~\text{ms}^{-1}\] \[a_c = \frac{(2(5)^2)^2}{100} = 25~\text{ms}^{-1}\] Now, calculate the magnitude of the total acceleration: \[a = \sqrt{(20)^2 + (25)^2} = \sqrt{625 + 1000} = \sqrt{1625} \approx 40.3~\text{ms}^{-1}\] #Conclusion# None of the given options are close to our calculated acceleration of approximately \(40.3~\text{ms}^{-1}\). Please check the problem statement or the provided options for possible errors.