Question

For an object whose velocity in ft/sec is given by v(t) = sin(t), what is its distance, in feet, travelled on the interval t = 1 to t = 5?
0.45
0.26
2.82
0.03

Answer 1

we have given velocity function v(t)=sint where t represent time .

we need to find distance traveled from to 1 to 5 time interval .

we need to use integration to find distance from velocity.

$\int\limits^a_b {v(t)} \, dt =distance traveled$.

we have a and b which is the time interval a=1 and b=5 .

$\int\limits^1_5 {v(t)} \, dt=\int\limits^1_5 {sint} \, dt=-cost =(-cos(5)-(-cos(1))$

we know that integration sint is - cost and then we need to plug the limits

distance traveled=$(-0.28-(-0.54)=0.26$