Final answer:
Using the quadratic formula, there are two values of t for which the equation a(t) = 0 is true: t=10 and t=-20. However, since negative time is typically not considered in these contexts, we only consider t=10 as the valid solution.
Step-by-step explanation:
To determine whether there are one or more values of t at which the equation a(t) = 0 is true, we can look at the provided quadratic equation t² + 10t - 200 = 0. To find the values of t, we use the quadratic formula which is t = (-b ± √(b² - 4ac)) / (2a).
To use the quadratic formula, we start by identifying the coefficients in the quadratic equation, which are a = 1, b = 10, and c = -200. We then substitute these values into the formula to find the values of t.
The determinant (the part under the square root in the formula) is b² - 4ac, which in this case is 10² - 4(1)(-200) or 100 + 800. Since this is a positive number, we know there will be two real solutions for t.
Substituting the coefficients into the quadratic formula gives us:
t = (-10 ± √(100 + 800)) / (2 × 1)
t = (-10 ± √900) / 2
t = (-10 ± 30) / 2
This yields two values for t:
- t = (20) / 2 which simplifies to t = 10
- t = (-40) / 2 which simplifies to t = -20
However, in the context of time, a negative value implies an event before the start of motion which usually is not considered. Therefore, we discard the negative value, leaving t = 10 as the solution.
Additionally, there is mention of finding the greatest instantaneous velocity, zero velocity, and negative velocity, but with the given information, we cannot solve these without additional context or equations relating to velocity.