The **question aims to find the acceleration of the cart** with initial speed **vo=5 m.s^(-1)**. The term **acceleration is defined as the rate of change of an object’s velocity with respect to time.** Accelerations are normally **vector quantities** (in that they have magnitude and direction). The **orientation of an object’s acceleration** is represented by the orientation of the **net force acting on that object**. The magnitude of the object’s acceleration, as described by **Newton’s second law**, is the combined effect of two causes:

**Net balance of all the external forces acting on that object**– the magnitude is directly proportional to that resulting resultant force;**Weight of that object**, depending on the materials it is made of- size is**inversely proportional**to the**object’s mass.**

The **system international units of acceleration is meter per second squared** $(m.s^{-2})$.

For example, when a **car starts from rest** (zero speed, in an inertial frame of reference) and travels in a straight line with increasing speed, it accelerates in the direction of travel. If the car turns, it will** accelerate in a new direction and change its motion vector.**

The **acceleration of the **car in its current direction of motion is termed **linear (or tangential in circular motions) acceleration**, the reaction to which is felt by the passengers on board as a force pushing them back into the seats of the car. When the direction is changing, the **applied acceleration is called radial** (or centripetal in circular motions) acceleration; the reaction passengers feel as **centrifugal force**.

**Expert Answer**

**Using the equation of motion equation:**

\[v^{2}=v_{o}^{2}+2ax\]

**For acceleration:**

\[a=\dfrac{v^{2}-v_{o}^{2}}{2x}\]

The **initial speed of the cart** is $v_{o}=5 m.s^{-1}$ at $x=0$, **reaches maximum displacement** at $x=12.5m$, at this petition, the cart starts to decelerate, the** velocity is zero** $v=0$ at this point because the **cart must stop for a moment before the cart changes its direction.**

**Plug the values to find the acceleration** as:

\[a=\dfrac{0-(5m.s^{-1})^{2}}{2(12.5m)}\]

\[=-1 m.s^{-2}\]

\[a=-1 m.s^{-2}\]

The **acceleration** is $-1 m.s^{-2}$.

**Numerical Result**

The **acceleration of the cart with the initial speed** $v_{0}=5 m.s^{-1}$ at position $x=0$ is given as $a=-1 m.s^{-2}$.

**Example**

The cart is powered by a large propeller or fan that can accelerate or decelerate the cart. The carriage starts at the position with an initial speed $v_{0}=10 m.s^{-1}$ and constant acceleration due to the fan. The direction to the right is positive. The carriage reaches the maximum position $x=15 m$, where it starts to move in the negative direction. Find the acceleration of the cart.

**Using the equation of motion equation:**

\[v^{2}=v_{o}^{2}+2ax\]

**For acceleration:**

\[a=\dfrac{v^{2}-v_{o}^{2}}{2x}\]

The **initial speed of the cart** is $v_{o}=10 m.s^{-1}$ at $x=0$, **reaches maximum displacement** at $x=15m$, at this petition, the cart starts to decelerate, the** velocity is zero** $v=0$ at this point because the **cart must stop for a moment before the cart changes its direction.**

**Plug the values to find the acceleration** as:

\[a=\dfrac{0-(10m.s^{-1})^{2}}{2(15m)}\]

\[=-3.33 m.s^{-2}\]

\[a=-3.33 m.s^{-2}\]

The **acceleration** is $-3.33 m.s^{-2}$.

The **acceleration of the cart with the initial speed** $v_{0}=10 m.s^{-1}$ at position $x=0$ is given as $a=-3.33 m.s^{-2}$.