A particle is moving on a straight line in such a way that its velocity v is given by v (t) = 2 t + 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? (Do not enter the units)

30

Step-by-step explanation:

Given the velocity of a particle to be;

V (t) = 2t+1

To get the distance travelled by the particle, we will use the formula

Velocity/speed = Distance/time

V = dS/dt

dS = Vdt ... 1

Substituting the value of the velocity function into equation 1, we have;

dS = (2t+1) dt

Integrating both sides to get the distance S, we have;

∫dS = ∫ (2t+1) dt

S = 2t²/2+t + C

S (t) = t²+t+C

To get the total distance travelled between the times t=0 and t=5, we will plug the value of t = 0 and t=5 into the function of the distance and then subtract as shown;

S (0) = 0²+0

S (0) = 0

Also

S (5) = 5²+5

S (5) = 25+5

S (5) = 30

Distance travelled between both times will be S (5) - S (0)

= 30-0

= 30

The total distance traveled by the particle is S = 30.

Step-by-step explanation:

Given that velocity,

v (t) = 2t + 1

To find the total distance travel, we integrate the velocity function, v (t), to obtain the distance function s (t), and evaluate the resulting distance at the interval given. That is at t = 0 to t = 5.

Integrating v (t) with respect to t, we have

s (t) = t² + t + C.

At t = 5

s (5) = 5² + 5 + C

= 25 + 5 + C

= 30 + C

At t = 0

S (0) = 0 + 0 + C

= C

The required distance is now

S (5) - S (0)

= 30 + C - C

= 30.