Chapter 2, Problem 76PQ

Two carts are set in motion at t = 0 on a frictionless track in a physics laboratory. The first cart is launched from an initial position of x = 18.0 cm with an initial velocity of 11.8î cm/s and a constant acceleration of –3.40î cm/s². The second cart is launched from x = 20.0 cm with a constant velocity of 4.30î cm/s.

1. a. N What are the times for which the two carts have equal speeds?
2. b. N What are the speeds of the carts at that time?
3. c. N What are the locations and times at which the carts pass each other?
4. d. C What is the difference between what is asked in parts (a) and (c) of this problem with regard to the times you found?

To determine

Time at which two carts moves with same speed.

Answer to Problem 76PQ

Time at which two carts moves with same speed will be $2.21\text{s}$.

Explanation of Solution

Write the Newton’s equation for velocity for the first cart.

    $v_1=u_1+a_{1}t$

Here, $v_1$ is the final velocity of first cart, $u_1$ is the initial velocity of first cart, $a_1$ is the acceleration of first cart, and $t$ is the time taken.

Write the Newton’s equation for velocity for the second cart.

    $v_2=u_2+a_{2}t$

Here, $v_2$ is the final velocity of second cart, $u_2$ is the initial velocity of second cart, and $a_2$ is the acceleration of second cart.

Equate the right hand sides’ of above equations.

    $u_1+a_{1}t=u_2+a_{2}t$

Rewrite the above equation in terms of $t$.

    $t=\frac{u_2-u_1}{a_1-a_2}$

Conclusion:

Substitute $4.30\text{cm}/\text{s}$ for $u_2$, $11.8\text{cm}/\text{s}$ for $u_1$, $-3.40\text{cm}/{\text{s}}^{\text{2}}$ for $a_2$, and $0\text{cm}/{\text{s}}^{\text{2}}$ in the above equation to find $t$.

  $\begin{array}{c}t=\frac{4.30\text{cm}/\text{s}-11.8\text{cm}/\text{s}}{-3.40\text{cm}/{\text{s}}^{\text{2}}-0\text{cm}/{\text{s}}^{\text{2}}}\\ =2.21\text{s}\end{array}$

Therefore, the time at which two carts moves with same speed will be $2.21\text{s}$.

To determine

Speed of carts at that time.

Answer to Problem 76PQ

Both carts will be at the speed of $4.30\text{cm}/\text{s}$.

Explanation of Solution

It is given that the first cart is only accelerating. Second one is moving at fixed velocity, $4.30\text{cm}/\text{s}$. So when both carts will be at equal speeds, means that the first cart also achieved the speed $4.30\text{cm}/\text{s}$.

Therefore, both carts will be at the speed of $4.30\text{cm}/\text{s}$.

To determine

The position and time when the carts pass each other.

Answer to Problem 76PQ

The time is $0.285\text{s}\text{and 4}\text{.13}\text{s}$ and the corresponding positions are $21.2\text{cm}\text{and 37}\text{.7}\text{cm}$ respectively.

Explanation of Solution

Write the Newton’s equation for displacement of first cart.

    $x_{1f}=x_{1i}+u_{1}t+\frac{1}{2}{a}_1{t}^2$

Here, $x_{1f}$ is the final position of first car and $x_{1i}$ is the initial position of car.

Write the Newton’s equation for displacement of first cart.

    $x_{2f}=x_{2i}+u_{2}t+\frac{1}{2}{a}_2{t}^2$

Here, $x_{2f}$ is the final position of second cart and $x_{2i}$ is the initial position of second cart.

Equate the right hand sides of above two equations.

  $x_{1i}+u_{1}t+\frac{1}{2}{a}_1{t}^2=x_{2i}+u_{2}t+\frac{1}{2}{a}_2{t}^2$

Conclusion:

Substitute $18.0\text{cm}$ for $x_{1i}$, $11.8\text{cm}/\text{s}$ for $u_1$, $-3.40\text{cm}/{\text{s}}^2$ for $a_1$, $20.0\text{cm}$ for $x_{2i}$, $0\text{cm}/{\text{s}}^2$ for $a_2$, $4.30\text{cm}/\text{s}$ for $u_2$ in the above equation to find $t$.

  $\begin{array}{c}18\text{cm}+\left(11.8\text{cm}/\text{s}\right)t+\frac{1}{2}\left(-3.40\text{cm}/{\text{s}}^2\right)t^2=20.0\text{cm}+\left(4.30\text{cm}/\text{s}\right)t+\frac{1}{2}\left(0\text{cm}/{\text{s}}^2\right)t^2\\ 18\text{cm}+\left(11.8\text{cm}/\text{s}\right)t-\frac{1}{2}\left(3.40\text{cm}/{\text{s}}^2\right)t^2=20.0\text{cm}+\left(4.30\text{cm}/\text{s}\right)t\\ 36\text{cm}+\left(23.6\text{cm}/\text{s}\right)t-\left(3.40\text{cm}/{\text{s}}^2\right)t^2=40.0\text{cm}+\left(8.60\text{cm}/\text{s}\right)t\\ \left(3.40\text{cm}/{\text{s}}^2\right)t^2-\left(15.0\text{cm}/\text{s}\right)t+4.0\text{cm}\text{=0}\end{array}$

Divide the above equation by $2$.

    $\left(1.70\text{cm}/{\text{s}}^2\right)t^2-\left(7.50\text{cm}/\text{s}\right)t+2.0\text{cm}\text{=0}$

Rewrite the above equation in terms of $t$

    $\begin{array}{c}t=\frac{\left(7.50\text{cm}/\text{s}\right)\pm \sqrt{{\left(7.50\text{cm}/\text{s}\right)}^2-4\left(1.70\text{cm}/{\text{s}}^2\right)2.0\text{cm}}}{2\left(1.70\text{cm}/{\text{s}}^2\right)}\\ =4.13\text{s}\text{or}\text{0}\text{.285}\text{s}\end{array}$

Substitute $18.0\text{cm}$ for $x_{1i}$, $11.8\text{cm}/\text{s}$ for $u_1$, $-3.40\text{cm}/{\text{s}}^2$ for $a_1$, and $\text{0}\text{.285}\text{s}$ for $t$ in the equation for $x_{1f}$ to find the position of cart at $\text{0}\text{.285}\text{s}$.

    $\begin{array}{c}{v}_{1f}=18\text{cm}+\left(11.8\text{cm}/\text{s}\right)\left(\text{0}\text{.285}\text{s}\right)+\frac{1}{2}\left(-3.40\text{cm}/{\text{s}}^2\right){\left(\text{0}\text{.285}\text{s}\right)}^2\\ =21.36\text{cm}-0.16\text{cm}\\ =21.2\text{cm}\end{array}$

Substitute $18.0\text{cm}$ for $x_{1i}$, $11.8\text{cm}/\text{s}$ for $u_1$, $-3.40\text{cm}/{\text{s}}^2$ for $a_1$, and $4.13\text{s}$ for $t$ in the equation for $x_{1f}$ to find the position of cart at $4.13\text{s}$.

  $\begin{array}{c}{v}_{1f}=18\text{cm}+\left(11.8\text{cm}/\text{s}\right)\left(4.13\text{s}\right)+\frac{1}{2}\left(-3.40\text{cm}/{\text{s}}^2\right){\left(4.13\text{s}\right)}^2\\ =66.73\text{cm}-29.03\text{cm}\\ =37.7\text{cm}\end{array}$

Therefore, the time is $0.285\text{s}\text{and 4}\text{.13}\text{s}$ and the corresponding positions are $21.2\text{cm}\text{and 37}\text{.7}\text{cm}$ respectively.

To determine

The difference in meaning of question in part (a) and part (b).

Answer to Problem 76PQ

Part (a) asks to find the instant at which carts having same speed and part (b) asks to identify the time at carts will be at the same position.

Explanation of Solution

In part (a), it is asked to find the time at which both carts having the same speed and in part (b), it is asked to find the time at which both carts are at the same location.

From the starting point to a moment just before $0.285\text{s}$, first car was moving behind the second car. But it was on acceleration. So at $0.285\text{s}$, first car passes the second car. At this point, both cars will at different velocity. After $0.285\text{s}$, first car starts decreasing its speed. So at $4.13\text{s}$, two cars came together once more. Between $0.285\text{s}$ and $4.13\text{s}$, at $2.21\text{s}$, velocity of cars became same.

Therefore, part (a) asks to find the instant at which carts having same speed and part (b) asks to identify the time at carts will be at the same position.