tes-rt-5t2-4i-9t-1j-and-writes-at-10i-9j-what-is-the-correct-acceleration-vector

Question

A student differentiates r→(t)=(5t2−4)i^+(9t−1)j^\overrightarrow{r}\left(t)\right.=\$$left(5t^2-4$$\right)\hat{i}+\left(9t-1\right)\hat{j} and writes a→(t)=10i^+9j^\overrightarrow{a}\left(t)\right.=10\hat{i}+9\hat{j}. What is the correct acceleration vector?

Accepted Answer

$$10i^10$$\hat{i}

Answer

5t2i^$$+9tj^5$t^2$$\hat{i}+9t\hat{j}

Answer

10ti^$$+9j^10t$$\hat{i}+9\hat{j}

Answer

10i^$$+9j^10$$\hat{i}+9\hat{j}