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An object, which is at the origin at time t=0, has initial velocity V0= (-14.0i - 7.0j)m/s and constant acceleration a=(6.0i + 3.0j)m/s^2..


Find the position x where the object comes to rest (momentarily).

Express your answer in terms of the unit vectors i and j.


r= _____m

Verified Answer :

antonsandiego

I am pretty sure the first thing we should do is to integrate acceleration into velocity
And we will have: \(a(t) = \ \textless \ 6,3\ \textgreater \\)\(v(t) = \ \textless \ (6t+C),(3t+C)\ \textgreater \ ; where -- v(0) = \ \textless \ -14,-7\ \textgreater \\)
Then calculate v :\(v(t) = \ \textless \ (6t-14),(3t-7)\ \textgreater \\)
As you can see, velocity is zero and none of the objects is moving. It happens when  t=7/3 which means we can calculate this in the way we did (integrating) :
\(p(t) = \ \textless \ (3t^2-14t+C) , ((3/2)t^2-7t+C) \ \textgreater \\)
The product is p(0) = 0 that makes us to reduce vector function to\(p(t) = \ \textless \ (3t^2-14t) , ((3/2)t^2-7t) \ \textgreater \ where t = 7/3\)\(p(7/3) = \ \textless \ -49/3 , -49/6 \ \textgreater \\)

johanrusli

The position x where the object comes to rest (momentarily) is

r = ( -16.3 i - 8.2 j ) m

Further explanation

Acceleration is rate of change of velocity.

\(\large {\boxed {a = \frac{v - u}{t} } }\)

\(\large {\boxed {d = \frac{v + u}{2}~t } }\)

a = acceleration (m / s²)v = final velocity (m / s)

u = initial velocity (m / s)

t = time taken (s)

d = distance (m)

Let us now tackle the problem!

This problem is about Kinematics.

Given:

vo = (-14.0i - 7.0j) m/s

a = (6.0i + 3.0j) m/s²

Unknown:

r = ? → v = 0 m/s

Solution:

To solve this problem, we need to use the following integral formula.

\(v = v_o + \int {a} \, dt\)

\(v = (-14.0 \, \widehat{i} - 7.0 \, \widehat{j}) + \int {(6.0 \, \widehat{i} + 3.0\, \widehat{j})} \, dt\)

\(v = (-14.0 + 6.0t) \, \widehat{i} + ( -7.0 + 3.0t ) \, \widehat{j}\)

If the object comes to rest (momentarily) , then :

\(v_x = 0\)

\((-14.0 + 6.0t) = 0\)

\(6.0t = 14\)

\(t = 14 \div 6.0\)

\(\boxed {t = \frac {7}{3} ~ s}\)

or

\(v_y = 0\)

\(( -7.0 + 3.0t ) = 0\)

\(3.0t = 7\)

\(t = 7 \div 3.0\)

\(\boxed {t = \frac {7}{3} ~ s}\)

\(r = r_o + \int {v} \, dt\)

\(r = 0 + \int { (-14.0 + 6.0t) \, \widehat{i} + ( -7.0 + 3.0t ) \, \widehat{j} } \, dt\)

\(r = ( -14.0t + 3.0t^2 ) \, \widehat{i} + ( -7.0t + 1.5t^2 ) \, \widehat{j}\)

At t = 7/3 s , then :

\(r = ( -14.0(\frac{7}{3}) + 3.0(\frac{7}{3})^2 ) \, \widehat{i} + ( -7.0(\frac{7}{3}) + 1.5(\frac{7}{3})^2 ) \, \widehat{j}\)

\(r = [ (-16\frac{1}{3}) \, \widehat{i} + (-8\frac{1}{6}) \, \widehat{j} ] ~ m\)

\(r \approx (-16.3 \, \widehat{i} - 8.2 \, \widehat{j} ) ~ m\)

Learn more

  • Velocity of Runner : https://brainly.com/question/3813437
  • Kinetic Energy : https://brainly.com/question/692781
  • Acceleration : https://brainly.com/question/2283922
  • The Speed of Car : https://brainly.com/question/568302

Answer details

Grade: High School

Subject: Mathematics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate , Sperm , Whale , Travel

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