Question

2 formula 1 point an object goes from 2.42 m/s to 42.10 m/s at 2.47 m/s/s. how long does it take? answer 3 formula 1 point an object is going 15.76 m/s after accelerating at 3.411 m/s/s for 7.25 sec. how fast was it going when it started? answer

Explanation:

Problem 2:

Step1: Recall the acceleration formula

The formula relating final velocity ($v$), initial velocity ($u$), acceleration ($a$), and time ($t$) is $v = u + at$. We need to find time $t$, so rearrange the formula to $t=\frac{v - u}{a}$.

Step2: Identify values

Given $u = 2.42\space m/s$, $v = 42.10\space m/s$, $a = 2.47\space m/s^2$.

Step3: Substitute values into the formula

$t=\frac{42.10 - 2.42}{2.47}=\frac{39.68}{2.47}$.

Step4: Calculate the result

$\frac{39.68}{2.47}\approx16.06\space s$.

Step1: Recall the acceleration formula

The formula is $v = u + at$, where $v$ is final velocity, $u$ is initial velocity, $a$ is acceleration, and $t$ is time. We need to find $u$, so rearrange the formula to $u=v - at$.

Step2: Identify values

Given $v = 15.76\space m/s$, $a = 3.411\space m/s^2$, $t = 7.25\space s$.

Step3: Substitute values into the formula

$u = 15.76-(3.411\times7.25)$.

Step4: Calculate the product

$3.411\times7.25 = 3.411\times(7 + 0.25)=3.411\times7+3.411\times0.25 = 23.877+0.85275 = 24.72975$.

Step5: Calculate the initial velocity

$u = 15.76 - 24.72975=-8.96975\space m/s$. The negative sign indicates the initial velocity is in the opposite direction of the final velocity (assuming the acceleration direction is positive).

Answer:

The time taken is approximately $\boldsymbol{16.06\space seconds}$.

Problem 3: