Science:Math Exam Resources/Courses/MATH110/December 2016/Question 06 (c)/Solution 1

By Hint, the acceleration is a derivative of the velocity;

$$a(t)=v'(t)=\left(15-{\frac {3}{4}}t\right)'=-{\frac {3}{4}}$$

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Plugging $t=0$, we get a $y$-intercept $y=-{\frac {3}{4}}$. On the other hand, since $a(t)$ only has its function value $-{\frac {3}{4}}$, there's no points whose function value is $0$. In other words, there's no $x$-intercepts.

Since it is a constant function, we can easily draw its graph as follows.