if the distance covered by an object in time is given by s(t)=t^2+5t where s(t) is in metersand t is in seconds, what is the distance covered in the interveral between 1 second and 5 seconds?
s(5)-s(1)
A 24 meters
B 30
C 40
D 42
E 44
so what would be the answer
plug in t=5 and t=1. subtract one from the other. what do you get? you've posted a lot of similar problems. By now you should be getting the hang of them.
30 ahh i get it now thanks
ummm. what did you do?
s(5) = 5^2 + 5*5 = 50
s(1) = 1^1 + 5*1 = 6
s(5)-s(1) = 44, not 30
i did it like this
5^2=25 + 5^1= 5 = 30
i guess i did it wrong
To find the distance covered in the interval between 1 second and 5 seconds, you need to evaluate the definite integral of the function s(t) = t^2 + 5t over that interval.
The definite integral of s(t) between 1 second and 5 seconds can be represented using the integral symbol (∫) and is calculated as follows:
∫[1 to 5] (t^2 + 5t) dt
To evaluate this integral, you can use the power rule of integration. Applying the power rule, you can integrate each term separately:
∫[1 to 5] t^2 dt + ∫[1 to 5] 5t dt
The antiderivative (integral) of t^2 with respect to t is (1/3) * t^3. The antiderivative of 5t is (5/2) * t^2.
Evaluating the integral for each term, you get:
[(1/3) * t^3] evaluated from 1 to 5 + [(5/2) * t^2] evaluated from 1 to 5
Substituting the upper and lower limits of integration, you get:
[(1/3) * (5^3) - (1/3) * (1^3)] + [(5/2) * (5^2) - (5/2) * (1^2)]
Simplifying the expression, you have:
[(125/3) - (1/3)] + [(125/2) - (5/2)]
Combining like terms, you get:
(124/3) + (120/2)
Converting the fractions to a common denominator of 6, you have:
(248/6) + (360/6)
Adding the fractions, you get:
(608/6)
Simplifying, the total distance covered between 1 second and 5 seconds is:
101.33 meters
that's 5t. not 5^t
5*5 = 25
you did it wrong in two ways. you tried to do 5^2 + 1^2 but did 5^2 + 5^1 instead.
You did not include both terms in your evaluations.
*sigh* review your algebra, if you want to do well in calculus.