Question

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

Answer 1

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$

Answer 2

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 : brainly.com/question/3813437

• Kinetic Energy : brainly.com/question/692781

• Acceleration : brainly.com/question/2283922

• The Speed of Car : 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