A Space vehicle accelerates uniformly from$\mathbf{85}\mathbf{}{\mathbf{ms}}^{\mathbf{-}\mathbf{1}}$at t = 0 to$\mathbf{162}\mathbf{}{\mathbf{ms}}^{\mathbf{-}\mathbf{1}}$at t = 10.0 s. How far did it move between t = 2.0 s and t = 6.0 s?

The rate at which velocity changes with respect to time is called acceleration. The SI unit of acceleration is${\mathrm{ms}}^{-2}$.

Velocity at t = 0 s,$v_{0\mathrm{s}}=85{\mathrm{ms}}^{-1}$

Velocity at t = 10.0 s, role="math" localid="1642845144797" $v_{10.0\mathrm{s}}=162{\mathrm{ms}}^{-1}$

Let the velocities at times t = 2.0 s and t = 6.0 s be $v_{2.0\mathrm{s}}$and $v_{6.0\mathrm{s}}$, respectively.

Also, let s represent the displacement in this duration.

Acceleration is defined as the rate of change of velocity.

The acceleration of the space vehicle is.

$\begin{array}{c}a=\frac{v_{10\mathrm{s}}-v_{0\mathrm{s}}}{10-0}\\ =\frac{162-85}{10}\\ =7.7{\mathrm{ms}}^{-2}\end{array}$

Using the first equation of motion, you get the velocity at t = 2.0 s as.

$v_{2.0\mathrm{s}}=v_{0\mathrm{s}}+a\left(2-0\right)$

Substituting the values in the above equation,

$\begin{array}{c}{v}_{2.0\mathrm{s}}=85+7.7\times 2\\ =100.4{\mathrm{ms}}^{-1}\end{array}$

Using the first equation of motion, you get the velocity at t = 6.0 s as.

$v_{6.0\mathrm{s}}=v_{0\mathrm{s}}+a\left(6-0\right)$

Substituting the values in the above equation,

$\begin{array}{c}{v}_{6.0\mathrm{s}}=85+7.7\times 6\\ =131.2{\mathrm{ms}}^{-1}\end{array}$

Using the third equation of motion, you get the distance traveled in the given time interval as.

$v_{6.0\mathrm{s}}^2-v_{2.0\mathrm{s}}^2=2as$

Here, s represents the displacement in the time interval 2.0 s to 6.0 s.

Substituting the values in the above equation,

$131.2^2-100.4^2=2\times 7.7\times s$

Solving the above equation for s,

$s=463.2\mathrm{m}$

Thus, the space vehicle will travel 463.2 m in the time interval 2.0 s to 6.0 s.