Question

FIRST ONE TO ANSWER GETS BRAINLIEST!!!!

The distance d in meters traveled by a skateboard on a ramp is related to the time t in seconds. This os modeled by the function: d(t) = 4.9t ^2 - 2.3t + 5. What is the maximum distance the skaeboard can travel, and at what time would it achieve this distance? Round your answer to the nearest hundredth.

Answer 1

Answer:

The maximum distance traveled is 4.73 meters in 0.23 seconds.

Step-by-step explanation:

We have that the distance traveled with respect to time is given by the function,

$d(t) = 4.9t^2-2.3t+5$.

Now, differentiating this function with respect to time 't', we get,

d'(t)=9.8t-2.3

Equating d'(t) by 0 gives,

9.8t - 2.3 = 0

i.e. 9.8t = 2.3

i.e. t = 0.23 seconds

Substitute this value in d'(t) gives,

d'(t) = 9.8 × 0.23 - 2.3

d'(t) = 2.254 - 2.3

d'(t) = -0.046.

As, d'(t) < 0, we get that the function has the maximum value at t = 0.23 seconds.

Thus, the maximum distance the skateboard can travel is given by,

$d(t) = 4.9\times 0.23^2-2.3\times 0.23+5$.

i.e. $d(t) = 4.9\times 0.0529-2.3\times 0.23+5$.

i.e. $d(t) = 0.25921-0.529+5$.

i.e. $d(t) = 0.25921-0.529+5$.

i.e. d(t) = 4.73021

Hence, the maximum distance traveled is 4.73 meters in 0.23 seconds.