A particle is moving on a straight line in such a way that its velocity v is given by v(t)=2t+1 for 0≤t≤5 where t is measured in seconds and v in meters per second. What is the total distance traveled (in meters) by the particle between times t=0 seconds and t=5 seconds ?
I was thinking I would do something like (2(0)+1) + (2(5)+1) but that didn't yield the correct answer which is 30.
if v(t) = 2t + 1
then s(t) = t^2 + t + c
s(5) = 25+5+c = 30+c
s)0) = c
effective distance = s(5) - s(0)
= 30+c - c
= 30
what you found is the difference in the velocities
Is there a way to do this without taking the integral of the equation which is what I believe you did? I'm just beginning integral calculus so we are still doing more basic methods.
To get distance from the velocity you must integrate.
To find the total distance traveled by the particle, we need to integrate the absolute value of the velocity function over the given time interval.
In this case, the velocity function is v(t) = 2t + 1. To find the absolute value of v(t), we can split it into two cases:
1. When 2t + 1 is positive or zero, the absolute value is equal to the original function: |v(t)| = 2t + 1.
2. When 2t + 1 is negative, the absolute value is the negation of the original function: |v(t)| = -(2t + 1).
Now, we need to identify the time periods during which the velocity is positive or zero, and when it is negative.
1. For 2t + 1 to be positive or zero, we solve the inequality:
2t + 1 ≥ 0
2t ≥ -1
t ≥ -1/2
Since the time interval is limited to 0 ≤ t ≤ 5, we can see that the velocity is always positive or zero within this interval.
2. Therefore, we only need to consider the case when 2t + 1 is negative.
When 2t + 1 < 0, we have:
2t + 1 < 0
2t < -1
t < -1/2
However, the given time interval is 0 ≤ t ≤ 5, which means there are no values of t that satisfy t < -1/2.
Hence, the particle's velocity is always positive or zero within the given time interval, and we can simplify the absolute value function as:
|v(t)| = 2t + 1
Now, to find the total distance traveled, we integrate |v(t)| over the given time interval:
∫[0,5] (2t + 1) dt
Integrating each term, we get:
∫[0,5] (2t) dt + ∫[0,5] (1) dt
Integrating both terms, we have:
[t^2] from 0 to 5 + [t] from 0 to 5
Substituting the upper and lower limits, we get:
(5^2) + (5) - (0^2) - (0)
Simplifying further, we have:
25 + 5 - 0 - 0 = 30
Therefore, the total distance traveled by the particle between t = 0 seconds and t = 5 seconds is 30 meters.