Question

At times=,${\mathbf{t}}_{\mathbf{1}}\mathbf{=}\mathbf{12}$s, a car is located at$<84,78,24>$ m and has velocity$<4,0,-3>$m/s. At time${\mathbf{t}}_{\mathbf{2}}\mathbf{=}\mathbf{18}$s, what is the position of the car? (The velocity is constant in magnitude and direction during this time interval.)

Short answer

The position of the car is: m.

Step-by-step solution

Identification of given data

Location at times,

Velocity at time s, =m/s

$$\begin{gathered} {\mathrm{t}}_1=12\mathrm{s} \\ {\mathrm{t}}_2=18\mathrm{s} \end{gathered}$$

Calculating position of the car

A position or position vector, also known as location vector or radius vector, is a Euclidean vector that represents the position of a point P in space in relation to an arbitrary reference originO.

To find the final position, use the following formula

This is valid because the velocity is constant and there is no acceleration.

Therefore,

$$\begin{gathered} \overrightarrow{v}=\frac{∆\overrightarrow{r}}{∆t} \\ \Rightarrow ∆\overrightarrow{r}=\overrightarrow{v}∆t \\ \Rightarrow \overrightarrow{r_2}=\overrightarrow{r_1}+\overrightarrow{v}\left(t_2-t_1\right) \end{gathered}$$

This is valid because the velocity is constant and there is no acceleration.

Therefore,

Thus, the position of the car is .