Calculus

Find the velocity vector for a particle whose position vector is vector components 2t, 5 for all t. (10 points)

A) vector components 2, 0
B) vector components t squared, 5 times t
C) vector components t cubed over 3, 5 times t squared over 2
D) the square root of the quantity 4 times t squared plus 25

  1. 👍
  2. 👎
  3. 👁
  1. r = <2t,5>
    r' = <2,0>

    1. 👍
    2. 👎
    👤
    oobleck

Respond to this Question

First Name

Your Response

Similar Questions

  1. Maths Vector Helpppp

    The points A and B have position vectors, relative to the origin O, given by −−→OA = i+2j+3k and −−→OB = 2i+j+3k. The line l has vector equation r = (1−2t)i+ (5+t)j+ (2−t)k. (i) Show that l does not intersect the

  2. calc

    If a particle moves in the xy-plane so that at time t>0 its position vector is (e^(t²), e^(-t³)), then its velocity vector at time t=3 is...? Thanks so much!

  3. Calculus and vectors

    Vector AB is a vector whose tail is at (-4,2) and whose head is at (-1,3). Calculate the magnitude of vector AB Determine the coordinates of point D on vector CD, if C (-6,0) and vector CD= vector AB. Please I need some help. Is

  4. Calculus

    Find the position vector of a particle whose acceleration vector is a = (6t, 2) with an initial velocity vector (0, 0) and initial position vector (3, 0) Please I need your help with explanation. Thank you

  1. Physics

    You are given vectors A = 5.0i - 6.5j & B = -3.5i + 7.0j. A third vector C lies on the xy-plane. Vector C is perpendicular to vector A, & the scalar product of C with B is 15.0. From this information, find the components of vector

  2. Calculus

    Find the magnitude of a particle's acceleration vector at t = 1 for a particle moving with position vector r(t)=(2t^2,4t^2-t^3) a) 4 b) 25 c) 5 d) 2√5

  3. Calculus

    Find the distance travelled from t = 0 to t = π of the point moving with position vector equation r equals vector components t cubed, 2 times the sine of t . (20 points) A) 8.239 B) 19.378 C) 25.693 D) 32.257

  4. Physics

    You are given vectors A = 5.0i - 6.5j and B = -2.5i + 7.0j. A third vector C lies in the xy-plane. Vector C is perpendicular to vector A and the scalar product of C with B is 15.0. Find the x and y components to vector C. Here's

  1. Calculus

    Find the position vector of a particle whose acceleration vector is a = (1, 2) with an initial velocity vector (0, 0) and initial position vector (1, 0). (10 points) Please give explanation and show work. Thank you so much!

  2. Physics

    You are given vectors A = 5.0i - 6.5j and B = -2.5i + 7.0j. A third vector C lies in the xy-plane. Vector C is perpendicular to vector A and the scalar product of C with B is 15.0. Find the x and y components to vector C. Answer

  3. Vectors

    Explain why it is not possible for Vector a • (Vector b • Vector c) to equal (Vector a • Vector b) • Vector c . (This means that the dot product is not associative.)

  4. Calculus

    What is the velocity vector for a moving particle with a position vector r(t)=(2/t, ln(t))? a) (2,e^t) b) (2ln(t), tln(t)) '/> c) (-2/t^2, 1/t) d) (ln(t), 2/t) My answer is "c" but is negative. Can you tell me if my answer is

You can view more similar questions or ask a new question.