Chapter 2, Problem 77GP

(a)

To determine

To Plot: The position vs time graph for both car’s start to the catchup point.

Explanation of Solution

Given:

Initial speed of police car, $u_p=0$

Constant speed of speeder, $u=120\text{ km/h}$

Formula used:

Second equation of kinematics:

  $S=ut+\frac{1}{2}a{t}^2\mathrm{..............}\text{(i)}$

Calculation:

Position vs time graph of speeder:

Speeder is moving with constant speed; therefore, it has zero acceleration.

From equation (i):

  $\begin{array}{c}S=ut+\frac{1}{2}\times 0\times t^2\\ =ut\\ S\propto t\end{array}$

Distance covered by speeder is directly proportional to the time. Therefore, the plot:

  

Position vs time graph of police car:

Police car start from the rest, to catch the speeder the police car needs to get accelerated.

From equation (i):

  $\begin{array}{c}S=0.t+\frac{1}{2}a{t}^2\\ S\propto t^2\end{array}$

Position of police car is directly proportional to the square of the time, therefore the plot between position and time would be parabola.

  

Conclusion:

Hence, the position vs time plot of speed would be straight line while the position vs time plot of police car would be parabola.

(b)

To determine

To Find: The time taken by police car to overtake the speeder.

Answer to Problem 77GP

  $t=22.5\text{ s}$

Explanation of Solution

Given:

Initial speed of police car, $u_p=0$

Constant speed of speeder, $u=120\text{ km/h = 120}\times \frac{5}{18}\text{ m/s = 33}\text{.33 m/s}$

Distance travelled, $S=750\text{ m}$

Acceleration of the speeder, $a=0$

Formula used:

Second equation of kinematics:

  $S=ut+\frac{1}{2}a{t}^2\mathrm{..............}\text{(i)}$

Calculation:

Let the time taken by the police car be $t$ .

From equation (i):

  $\begin{array}{c}750=33.33t+\frac{1}{2}\times 0\times t^2\\ t=\frac{750}{33.33}\\ =22.5\text{ s}\end{array}$

Conclusion:

Time taken by the police car to overtake the speeder: $22.5\text{ s}$

(c)

To determine

To Find: The acceleration of the police car.

Answer to Problem 77GP

  $a_p=2.96{\text{ m/s}}^2$

Explanation of Solution

Given:

Initial speed of police car, $u_p=0$

Distance travelled by the police car, $S=750\text{ m}$

Time taken to catch the speeder, $t=22.5\text{ s}$

Formula used:

Second equation of kinematics:

  $S=u_{p}t+\frac{1}{2}{a}_p{t}^2$

Where,

  $S=$ Distance travelled by police car.

  $u_p=$ Initial speed of police car.

  $a_p=$ Acceleration of police car.

Calculation:

Let the acceleration of the police car be $a_p$ .

Plugging in the given values in formula:

  $\begin{array}{c}750=0\times 22.5+\frac{1}{2}{a}_p\times {\left(22.5\right)}^2\\ a_p=\frac{2\times 750}{{\left(22.5\right)}^2}\\ =2.96{\text{ m/s}}^2\end{array}$

Conclusion:

Acceleration of the police car: $2.96{\text{ m/s}}^2$

(d)

To determine

To Find: The speed of police car at the overtaking point.

Answer to Problem 77GP

  $v_p=66.6\text{ m/s}$

Explanation of Solution

Given:

Initial speed of police car, $u_p=0$

Time taken to catch the speeder, $t=22.5\text{ s}$

Acceleration of the police car, $a_p=2.96{\text{ m/s}}^2$

Formula used:

First equation of kinematics:

  $v_p=u_p+a_{p}t$

Where,

  $u_p=$ Initial speed of police car.

  $a_p=$ Acceleration of police car.

  $v_p=$ Speed of police car after time $t$ .

Calculation:

Plugging in the given values in the formula:

  $\begin{array}{c}{v}_p=0+2.96\times 22.5\\ =66.6\text{ m/s}\end{array}$

Conclusion:

Speed of car at the overtaking point: $66.6\text{ m/s}$