Question

The crew of a stationary spacecraft observe an asteroid whose mass is$\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{17}}$kg. Taking the location of the spacecraft as the origin, the asteroid is observed to be at location$<-3\times {10}^3,-4\times {10}^3,8\times {10}^3>$m at a time $\mathbf{18}\mathbf{.}\mathbf{4}$s after lunchtime. At a time $\mathbf{21}\mathbf{.}\mathbf{4}$s after lunchtime, the asteroid is observed to be at location$<-1.4\times {10}^3,-6.2\times {10}^3,9.7\times {10}^3>$m. Assuming that the velocity of the asteroid does not change during this time interval, calculate the vector velocityrole="math" localid="1656672441674" $\overrightarrow{\mathbf{v}}$of the asteroid.

Short answer

The velocity of the asteroid is:m/s.

Step-by-step solution

Identification of given data

Mass-$4\times {10}^{17}$kg

Location at a time$18.4$s after lunchtime-m

Location at a time s after lunchtime- m

Calculating velocity

The velocity of the asteroid is the change in position vector over the change in time, that is

Hence, the velocity vector is: .