Question

A particle's position as a function of time t is given by r⃗ =(5.0t+6.0t2)mi^+(7.0−3.0t3)mj^.

Answer 1

The angle of the particle's displacement vector is -65° from the positive x axis

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 !

Given :

$\overrightarrow{r_1} = ( 5.0 ~ t + 6.0 ~ t^2) \widehat{i} + ( 7.0 - 3.0 ~ t^3 ) \widehat{j}$

$\overrightarrow{r_2} = ( 0.0 )\widehat{i} + ( 7.0 ) \widehat{j}$

The find the displacement of the particles is to subtract the two position vectors:

$\overrightarrow{\Delta r} = \overrightarrow{r_1} - \overrightarrow{r_2}$

$\overrightarrow{\Delta r} = [( 5.0 ~ t + 6.0 ~ t^2) \widehat{i} + ( 7.0 - 3.0 ~ t^3 ) \widehat{j}] - [( 0.0 )\widehat{i} + ( 7.0 ) \widehat{j}]$

$\overrightarrow{\Delta r} = ( 5.0 ~ t + 6.0 ~ t^2) \widehat{i} + ( - 3.0 ~ t^3 ) \widehat{j}$

At t = 5.0 s ,

$\overrightarrow{\Delta r} = ( 5.0 ~ t + 6.0 ~ t^2) \widehat{i} + ( - 3.0 ~ t^3 ) \widehat{j}$

$\overrightarrow{\Delta r} = ( 5.0 ~ (5.0) + 6.0 ~ (5.0)^2) \widehat{i} + ( - 3.0 ~ (5.0)^3 ) \widehat{j}$

$\overrightarrow{\Delta r} = 175 \widehat{i} - 375 \widehat{j}$

$\tan \theta = \frac{-375}{175}$

$\tan \theta = \frac{-15}{7}$

$\theta = \tan^{-1} (\frac{-15}{7})$

$\large {\boxed {\theta \approx -65^o} }$

Learn more

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• The Speed of Car : brainly.com/question/568302

Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

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

Answer 2

The given particle position is
$\vec{r}(t)=(5.0t+6.0t^{2})\hat{i}+(7.0-3.0t^{3})\hat{j}$

When t=5 s, the position vector is
$\vec{r}_{1} = (5+150)\hat{i}+(7-375)\hat{j} = 175\hat{i}-368\hat{j}$

The reference vector at t = 5 s is
$\vec{r}_(2) = 0\hat{i}+7\hat{j}.$

Let θ = the angle between the two vectors.
Then, by definition,
$\vec{r}_{1}.\vec{r}_{2}=|r_{1}||r_{2}|cos\theta$

|r₁} =  √[175²+(-368)²] = 407.4911
|r₂| = 7
$\vec{r}_{1}.\vec{r}_{2}=(-368)*(7) = -2576$

Therefore
θ = cos⁻¹ -2576/(407.4911*7) = cos⁻¹ -0.9031 = 154.57°

Answer: 154.57°