Question

A cart is propelled over an XY plane with acceleration components $$a_x=4m/s^2$$ and$$a_y=-2.0m/s^2$$. Its initial velocity has components $$v_{0x}=8m/s$$ and $$v_{0y}=12m/s$$. In unit-vector notation, what is the velocity of the cart when it reaches its greatest y coordinate?

Answer 1

Hint: In this solution, we will use the general equations of kinematics to determine the velocity of the cart. When the cart is at its greatest y coordinate it will have zero velocity in the y-direction.
Formula used: In this solution, we will use the following formula:
First equation of kinematics: $v=u+at$ where $v$ is the final velocity of the cart, $u$ is the initial velocity $u$ under acceleration $a$ in time $t$

Complete step by step answer:
We want to find the velocity of the cart when the cart is at its greatest coordinate. To do that, first, we will find the time the cart takes to reach its greatest y coordinate knowing that the velocity of the cart at its highest y-coordinate will be zero as the object is retarding in the y-direction (since its acceleration value is negative).
So, using the first law of kinematics, we can write
$v_y=u_y+at$
Since $v_y=0$, and $u_y=12m/s$ and $$a_y=-2.0m/s^2$$, we can write
$0=12-2t$
Which gives us
$t=6s$.
This is the time taken by the cart to reach the point where its velocity in the y-direction will be zero. Since the velocity of the cart in the y-direction will be zero at its highest point, its net velocity will be due to the velocity in the x-direction. Given that we’ve found out the time taken by the object to reach its greatest y coordinate, we can then use the velocity gained by the object in the x-direction using the first law of kinematics again as
$v_x=u_x+a_{x}t$
As $u_x=8m/s$ and $a_x=4m/s^2$, we can write
$v_x=8+4(6)$
$\Rightarrow v_x=32\hat{i}m/s$. Which will be the velocity of the cart when it is at its highest point. Here $\hat{i}$ denotes that the direction of the object will be in the x-direction.

Note: We must realize that the velocity of the cart when it is at its highest point will be purely in x-direction as the velocity in the y-direction will be zero. This motion is similar to projectile motion however the difference is that the object is being accelerated in x-direction which is not the case for a projectile motion.