Question

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acceleration (m/s²)
acceleration of cart
force (n)
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for the graph, acceleration of cart, the acceleration of the cart when the
applied force was 1.2 n is

m/s².

Explanation:

Step1: Find two points on the line

Let's take two clear points. For example, when force \(F_1 = 0.2N\), acceleration \(a_1=2.0m/s^{2}\); when \(F_2 = 1.6N\), \(a_2 = 0.5m/s^{2}\).
The equation of a line in the form \(y = mx + c\) (here \(a= mF + c\)). The slope \(m=\frac{a_2 - a_1}{F_2 - F_1}=\frac{0.5 - 2.0}{1.6 - 0.2}=\frac{- 1.5}{1.4}\approx - 1.07\). Using the point - slope form \(a - a_1=m(F - F_1)\), \(a-2.0=-1.07(F - 0.2)\), \(a=-1.07F+2.0 + 0.214\), \(a=-1.07F + 2.214\).

Step2: Substitute \(F = 1.2N\) into the equation

When \(F = 1.2N\), \(a=-1.07\times1.2 + 2.214\).
\(a=-1.284+2.214\).

Answer:

\(0.93m/s^{2}\)