Question

On release 1 a student releases a cart from the top of a ramp such that the cart was travel 200 cm before hitting the stopper. The cart increase the speed steadily and reaches the bottom of the ramp in 4.40 seconds. What was the magnitude of the cart acceleration on release 1 ? The student then releases the cart from a lower position such that the cart only travels 70.0 cm. What was the magnitude of the carts acceleration on release 2? How long did the car take to reach the bottom of the ramp on release 2? Please see the picture

Answer 1

0. a = 20.66 cm/s^2

,

1. a = 20.66 cm/s^2

,

2. t = 2.60 s

Step-by-step explanation

1. The initial speed in release 1 is zero. The distance the cart travels is x = 200cm

Which we can write as a function of time:

`x=x_0+v_0t+(1)/(2)at^2`

We know that the cart traveled for 4.40 seconds. Replacing with the information we have:

`\begin{gathered} 200cm=0+0\cdot4.40s+(1)/(2)a\cdot4.40^2s^2 \\ 200cm=(1)/(2)\cdot a\cdot4.40^2s^2 \end{gathered}`

Solving for a:

`a=(2\cdot200cm)/(4.40^2s^2)=20.66cm/s^2`

The magnitude of the cart's acceleration is 20.66 cm/s^2, rounded tot he nearest hundredth.

2. Now the students wants the cart to travel x = 70 cm instead. If it is the same cart and the same ramp, with the same slope, then the forces acting upon the cart are the same. This means that it doesn't matter from what position of the ramp the cart is released, it will have the same acceleration every time. However in this case, it will travel for a shorter period of time.

Hence the acceleration in release 2 is the same as in release 2: 20.66 cm/s^2

3. In this case we know that x = 70cm, a = 20.66 cm/s^2 and we have to find t. From this formula:

`x=x_0+v_0t+(1)/(2)at^2`

Replacing with the information we have:

And solving for t:

We find that the cart traveled for 2.60 seconds, rounded to the nearest hundredth.