91Ó°ÊÓ

Acceleration: ${\text{a}}_1=2.25\text{m}/{\text{s}}^2$

Acceleration: ${\text{a}}_2=-0.750\text{m}/{\text{s}}^2$

Displacement: $x=900\text{m}$

Displacement: $x_1=\frac{900}{4}=225\text{m}$

Displacement: $x_2=\frac{3\left(900\right)}{4}\text{m}$

The problem deals with the kinematic equation of motion in which the motion of an object is described at constant acceleration.Using the second kinematic equation, the time to travel the distance 900 m can be found. Using the formula for the third kinematic equation, find the maximum velocity of the car.

Formula:

The displacement in kinematic equations is given by,

$x=v_{0}t+\frac{1}{2}a{t}^2$

The final velocity is given by,

$v_f^2=v_0^2+2ax$

A car travels distance of 900 m along x axis and starts at rest $(x=0m)$ and ends at rest $(x=900\mathrm{m})$.

For first $\frac{1}{4}$of that distance, acceleration is $+2.25{\text{m/s}}^{\text{2}}$ and for rest of that distance, it’s acceleration is $-0.750{\text{m/s}}^{\text{2}}$.

We have,

$x_1=v_{01}{t}_1+\frac{1}{2}{a}_1{t_1}^2$ (i)

Where,

$\begin{array}{rcl}{a}_1& =& +2.25{\text{m/s}}^{\text{2}}\\ x_1& =& \frac{900}{4}\text{m}\\ v_{01}& =& 0\text{m/s}\end{array}$

$t_1=14.14\text{s}$

$x_2=v_{02}{t}_2-\frac{1}{2}{a}_2{t_2}^2$ (ii)

Where,

$\begin{array}{rcl}{a}_2& =& -0.75{\text{m/s}}^{\text{2}}\\ x_2& =& \frac{3\left(900\right)}{4}\text{m}\\ v_{02}& =& 0\text{m/s}\end{array}$

$t_2=42.46s$

So, the total time is $t=t_1+t_2=14.14+42.46=56.6s$.

$\begin{array}{rcl}{v}^2& =& (v_{01}{)}^2+2a_1{x}_1\\ v^2& =& 0+2\left(2.25{\text{m/s}}^{\text{2}}\right)\left(225\text{m}\right)\\ & =& 1013{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ v& =& 31.8\text{m/s}\end{array}$

So the maximum speed is $v=31.8\text{m/s}$.