Question

a high-speed particle’s acceleration, in \\(\frac{m}{s^2}\\), at a particular time \\(t\\) is given by the function \\(a(t) = t^4 + 9t^3 - 524\\). what is the particle’s acceleration, in \\(\frac{m}{s^2}\\), at \\(t = 5.9\\) seconds? round to one decimal place.
show your work here
hint: to add an exponent (\\(x^2\\)), type \exponent\ or press \^\

Explanation:

Step1: Substitute t = 5.9 into the function

We have the acceleration function \( a(t) = t^4 + 9t^3 - 524 \). Substitute \( t = 5.9 \) into the function:
\( a(5.9) = (5.9)^4 + 9\times(5.9)^3 - 524 \)

Step2: Calculate each term

First, calculate \( (5.9)^4 \):
\( 5.9^4 = 5.9\times5.9\times5.9\times5.9 = 34.81\times5.9\times5.9 = 205.379\times5.9 = 1211.7361 \)

Next, calculate \( 9\times(5.9)^3 \):
\( 5.9^3 = 5.9\times5.9\times5.9 = 34.81\times5.9 = 205.379 \)
\( 9\times205.379 = 1848.411 \)

Step3: Combine the terms and subtract 524

Now, add the two calculated terms and subtract 524:
\( a(5.9) = 1211.7361 + 1848.411 - 524 \)
First, add 1211.7361 and 1848.411: \( 1211.7361 + 1848.411 = 3060.1471 \)
Then subtract 524: \( 3060.1471 - 524 = 2536.1471 \)

Step4: Round to one decimal place

Round 2536.1471 to one decimal place. The second decimal digit is 4, which is less than 5, so we keep the first decimal digit as is. So, 2536.1 (rounded to one decimal place). Wait, wait, no, wait. Wait, did I make a mistake? Wait, let's recalculate:

Wait, \( 5.9^4 \): Let's compute 5.9^2 = 34.81, 5.9^3 = 34.815.9 = 34.816 - 34.810.1 = 208.86 - 3.481 = 205.379, 5.9^4 = 205.3795.9. Let's compute 205.3796 = 1232.274, minus 205.3790.1 = 20.5379, so 1232.274 - 20.5379 = 1211.7361. That's correct.

9(5.9)^3: 5.9^3 is 205.379, 9205.379 = 1848.411. Correct.

Then 1211.7361 + 1848.411 = 3060.1471. Then 3060.1471 - 524 = 2536.1471. Now, rounding to one decimal place: the number is 2536.1471. The first decimal is 1, the second is 4, so we round down, so 2536.1? Wait, but that seems high. Wait, maybe I made a mistake in the function. Wait, the function is \( a(t) = t^4 + 9t^3 - 524 \). Let's check with t=6: t=6, t^4=1296, 9t^3=9*216=1944, so 1296+1944=3240, 3240-524=2716. So at t=5.9, which is less than 6, the value should be less than 2716. Our calculation gave 2536.1, which is less than 2716, so that's reasonable. Wait, but let's check the calculation again.

Wait, 5.9^4: 5.95.9=34.81; 34.815.9=34.81(6-0.1)=34.816 - 34.810.1=208.86 - 3.481=205.379; 205.3795.9=205.379(6-0.1)=205.3796 - 205.379*0.1=1232.274 - 20.5379=1211.7361. Correct.

9(5.9)^3: 5.9^3=205.379, 9205.379=1848.411. Correct.

1211.7361 + 1848.411=3060.1471. Then 3060.1471 - 524=2536.1471. Rounding to one decimal place: look at the second decimal digit, which is 4, so we round down, so 2536.1. Wait, but let's use a calculator for more accuracy. Let's compute 5.9^4: 5.9^4 = (5.9^2)^2 = (34.81)^2 = 34.8134.81. Let's compute that: 3434=1156, 340.81=27.54, 0.8134=27.54, 0.810.81=0.6561. So (34 + 0.81)^2 = 34^2 + 2340.81 + 0.81^2 = 1156 + 55.08 + 0.6561 = 1211.7361. Correct. So 5.9^4=1211.7361. 95.9^3=9(5.95.95.9)=9(34.815.9)=9205.379=1848.411. Then 1211.7361 + 1848.411=3060.1471. 3060.1471 - 524=2536.1471. Rounded to one decimal place: 2536.1. Wait, but maybe the original function was \( t^4 + 9t^2 - 524 \)? No, the problem says \( 9t^3 \). So I think the calculation is correct.

Answer:

2536.1