Q88P - Fundamentals Of Physics

Question

A car moving with constant acceleration covered the distance between two points $\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{m}$ apart in $\mathbf{6}\mathbf{.}\mathbf{00}\mathit{s}$. Its speed as it passed the second point was $\mathbf{15}\mathbf{.}\mathbf{0}\mathit{m}\mathbf{/}\mathit{s}$. (a) What was the speed at the first point? (b) What was the magnitude of the acceleration? (c) At what prior distance from the first point was the car at rest? (d) Graph x versus t and v  versus t for the car, from rest $(t =0)$.

Step-by-Step Solution

Answer

1. The speed of the car at the first point is $5 \mathrm{m}/\mathrm{s}$
2. The magnitude of the acceleration is $1.67 \mathrm{m}/{\mathrm{s}}^2$
3. The prior distance from the first point where the car was at rest is $-7.5 \mathrm{m}$
4. The graph of  $\mathrm{x} \mathrm{vs} \mathrm{t} \mathrm{and} \mathrm{v} \mathrm{vs} \mathrm{t}$

Step 1: Given information

$\begin{array}{rcl}\Delta x& =& 60 \mathrm{m}\\ {\mathrm{v}}_{\mathrm{f}}& =& 15 \mathrm{m}/\mathrm{s}\\ \mathrm{t}& =& 6.0 \mathrm{s}\end{array}$

Step 2: To understand the concept

This problem deals with the kinematic equation of motion. Kinematics is the studies of how a system of bodies moves without taking into account the forces or potential fields that influence the motion. The equations which are used in the study are known as kinematic equations of motion.

Formula:

The final velocity in the kinematic equation is expressed as,

$\begin{array}{rcl}{v}_f& =& v_0+at.......\left(i\right)\\ {v_f}^2 & =& {v_0}^2 + 2a \left(\Delta x\right) .......\left(ii\right)\end{array}$

Step 3 (a): Calculations for speed of the car at first point

According to equation (i), the velocity is

 $\begin{array}{rcl}15& =& v_{0 }+a\left(6\right)\\ a& =& \frac{15-v_{0 }}{6}\end{array}$

Substituting a in equation (ii),

$\begin{array}{rcl}{v_f}^2 & =& {v_0}^2 + 2a \left(\Delta x\right) \\ 225& =& {v_0}^2+2\left(\frac{15- v_{0 }}{6}\right)60 \\ 225& =& {v_0}^2+300-20 v_{0 }\\ {v_0}^2-20v_{0 }+75& =& 0 \\ v_{0 }& =& 5 \mathrm{m}/\mathrm{s}\end{array}$ 

We get the initial velocity of the car as $5 \mathrm{m}/\mathrm{s}$.

Step 4 (b): Calculations for acceleration

From part (a), the acceleration is, $a=\frac{15- v_{0 }}{6}$

 $\begin{array}{c}a=\frac{15-5}{6}\\ =1.67 \mathrm{m}/{\mathrm{s}}^2\end{array}$

So, the acceleration of car would be $1.67 \mathrm{m}/{\mathrm{s}}^2.$

Step 5 (c): Calculation of the distance until it comes to rest

To find the distance from the first point where the car was at rest we have,

$\begin{array}{rcl} v_{f }& =& 0 \mathrm{m}/\mathrm{s}\\ {\mathrm{v}}_{0 }& =& 5 \mathrm{m}/\mathrm{s} \\ \mathrm{a} & =& 1.67 \mathrm{m}/{\mathrm{s}}^2\end{array}$ 

Using equation (ii)

 $\begin{array}{rcl}{v_f}^2 & =& {v_0}^2 + 2a \left(\Delta x\right) \\ (0{)}^2 & =& (5{)}^2 + 2\left(1.67\right) \left(\Delta x\right)\\ \left(\Delta x\right)& =& -7.5 \mathrm{m}.\end{array}$

So, the distance traveled before coming to rest would be $7.5 \mathrm{m}.$According to kinematic equations,

$\begin{array}{rcl}t& =& \frac{v_f- v_0}{a}\\ t & =& \frac{0 -- 5}{1.67}\\ t& =& -3\sec .\end{array}$

 

Step 6 (d): Graphical representation