Q36P - Fundamentals Of Physics

Question

A car moves along an x axis through a distance of 900 m, starting at rest (at $\mathit{x}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{0}$) and ending at rest (at $\mathit{x}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{900}\mathbf{ }\mathbf{m}$). Through the first $\frac{\mathbf{1}}{\mathbf{4}}$ of that distance, its acceleration is  $\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{25}\mathbf{ }\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$. Through the rest of that distance, its acceleration is  $\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{750}\mathbf{ }\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$. What are (a) its travel time through 900 m. (b) its maximum speed? (c) Graph position x, velocity v and acceleration a versus time t for the trip.

Step-by-Step Solution

Answer

(a) Travel time of the car through the 900 m is $56.6 \text{s}$.

(b) The maximum speed of the car is  $31.8 \text{m/s}$.

Given information

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

Understanding the concept

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) Determination of travel time for a car for 900 m

A car travels distance of 900 m along x axis and starts at rest $(x=0 m)$  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  $t=t_1+t_2=14.14+42.46=56.6s$.

(b) Determination of maximum speed of car

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

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

(c) Graph of displacement vs time

(c) Graph of velocity vs time

(c) Graph of acceleration vs time