Question

3.2. A quadbike is originally at the point with position vector, \vec{r} = (-2.53\hat{i} + 28.9\hat{j}) m. moving with initial velocity, \vec{v} = (3.45\hat{i} + 2.89\hat{j}) m/s. It moves with constant acceleration, \vec{a} = (-1.79\hat{i} - 0.650\hat{j}) m/s^2. After 5 seconds have elapsed, find (a) Its velocity vector and speed. (b) Its position vector.

          3.2. A quadbike is originally at the point with position vector, \vec{r} = (-2.53\hat{i} + 28.9\hat{j}) m. moving with initial velocity, \vec{v} = (3.45\hat{i} + 2.89\hat{j}) m/s. It moves with constant acceleration, \vec{a} = (-1.79\hat{i} - 0.650\hat{j}) m/s^2. After 5 seconds have elapsed, find
(a) Its velocity vector and speed.
(b) Its position vector.

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3.2. A quadbike is originally at the point with position vector, r⃗ = (-2.53î + 28.9ĵ) m. moving with initial velocity, v⃗ = (3.45î + 2.89ĵ) m/s. It moves with constant acceleration, a⃗ = (-1.79î - 0.650ĵ) m/s^2. After 5 seconds have elapsed, find
(a) Its velocity vector and speed.
(b) Its position vector.

Added by Katie G.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Adi S

12/03/2023

Step 1

Given: Initial velocity, u = 3.45i + 2.89j m/s Acceleration, a = -1.79i - 0.650j m/s^2 Time elapsed, t = 5 seconds Using the equation v = u + at: v = (3.45i + 2.89j) + (-1.79i - 0.650j) * 5 v = 3.45i + 2.89j - 8.95i - 3.25j v = -5.50i - 0.36j Therefore, the Show more…

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3.2. A quadbike is originally at the point with position vector, r = (-2.53i + 28.9j) m, moving with initial velocity, v = (3.45i + 2.89j) m/s. It moves with constant acceleration, a = (-1.79i - 0.650j) m/s^2. After 5 seconds have elapsed, find: (a) Its velocity vector and speed. (b) Its position vector.

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