Question

how far does a toy car travel in 6.0 s if it is pushed down a ramp at an initial speed of 2.5 m/s and has an acceleration of 2.7 m/s²?

Explanation:

Step1: Identify the kinematic equation

The kinematic equation \(x = v_0t+\frac{1}{2}at^{2}\) is used, where \(v_0\) is the initial velocity, \(t\) is the time, and \(a\) is the acceleration.

Step2: Substitute the given values

Given \(v_0 = 2.5\ m/s\), \(t = 6.0\ s\), and \(a=2.7\ m/s^{2}\).
Substitute into the equation:
\(x=(2.5\times6.0)+\frac{1}{2}(2.7)(6.0)^{2}\)
First calculate \(2.5\times6.0 = 15\)
Then calculate \(\frac{1}{2}(2.7)(6.0)^{2}=\frac{1}{2}(2.7)\times36 = 48.6\)

Step3: Calculate the total distance

\(x=15 + 48.6\)

Answer:

\(63.6\ m\)