on-at-6t-4i-3t2j-for-1-t-3-s-what-is-the-change-in-velocity-v-over-this-interval

Question

A puck on an air table has acceleration a→(t)=(6t−4)i^+(3t2)j^\overrightarrow{a}\left(t)\right.=\left(6t-4\right)\hat{i}+\$$left(3t^2$$\right)\hat{j} for 1≤t≤31 ≤ t ≤ 3 s. What is the change in velocity ΔvΔv over this interval?

Accepted Answer

16i^$$+26j^16$$\hat{i}+26\hat{j}

Answer

15i^$$+26j^15$$\hat{i}+26\hat{j}

Answer

16i^$$+27j^16$$\hat{i}+27\hat{j}

Answer

14i^$$+28j^14$$\hat{i}+28\hat{j}

A puck on an air table has acceleration a→(t)=(6t−4)i^+(3t2)j^\(\overrightarrow{a}\)\(\left\)(t)\(\right\).=\(\left\)(6t-4\(\right\))\(\hat{i}\)+\(\left\)(3t^2\(\right\))\(\hat{j}\) for 1≤t≤31 ≤ t ≤ 3 s. What is the change in velocity ΔvΔv over this interval?

A puck on an air table has acceleration $\vec{a} (t) = (6 t - 4) \hat{i} + (3 t^{2}) \hat{j}$ for $1 is less than or equal to t is less than or equal to 3$ s. What is the change in velocity $Δ v$ over this interval?