Question

The velocity function (in meters per second) is given for a particle moving along a line. v(t) = 5t − 9, 0 ≤ t ≤ 3 (a) Find the displacement. -9/2 Correct: Your answer is correct. m (b) Find the distance traveled by the particle during the given time interval.

Answer 1

Answer:

a) The displacement is -4.5 m.

b) The traveled distance is 11.7 m.

Explanation:

Hi there!

a)The velocity of the particle is the derivative of the displacement function, x(t):

v(t) = dx/dt = 5t - 9

Separating varibles:

dx = (5t - 9)dt

Integrating both sides from x = x0 to x and from t = 0 to t.

x - x0 = 1/2 · 5t² - 9t

x = 1/2 · 5t² - 9t + x0

If we place the origin of the system of reference at x = x0, the displacement at t = 3 will be x(3):

x(3) = 1/2 · 5 · (3)² - 9(3) + 0

x(3) = -4.5

The displacement at t = 3 s is -4.5 m. It means that the particle is located 4.5 m to the left from the origin of the system of reference at t = 3 s.

b) When the velocity is negative, the particle moves to the left. Let´s find the time at which the velocity is less than zero:

v = 5t - 9

0 > 5t - 9

9/5 > t

1.8 s > t

Then until t = 1.8 s, the particle moves to the left from the origin of the reference system.

Let´s find the position of the particle at that time:

x = 1/2 · 5t² - 9t

x = 1/2 · 5(1.8 s)² - 9(1.8 s)

x = -8.1 m

From t = 0 to t = 1.8 s the traveled distance is 8.1 m. After 1.8 s, the particle has positive velocity. It means that the particle is moving to the right, towards the origin. If at t = 3 the position of the particle is -4.5 m, then the traveled distance from x = -8.1 m to x = -4.5 m is (8.1 m - 4.5 m) 3.6 m.

Then, the total traveled distance is (8.1 m + 3.6 m) 11.7 m.

Answer 2

Answer:

The displacement is -4.5 m.

The distance covered is 15.3 m.

Explanation:

The velocity function is, $v(t)=5t-9$. at $0\leq t\leq 3$.

(a)

The displacement (x) is,

$v(t)=\dfrac{dx}{dt}\\ dx=\int\limits^3_0 {v(t)} \, dt\\ dx=\int\limits^3_0 {(5t-9)} \, dt$

Integrating above expression as,

$x=\dfrac{5}{2}t^{2}-9t \\x=\dfrac{5}{2}(3)^{2}-9(3)\\x= -4.5 \;\rm m$

Thus, the displacement is -4.5 m. (Negative sign shows that displacement is at left of origin)

(b)

At initial, the particle is at rest (v=0). Then time taken is,

$v(t)=5t-9\\0=5t-9\\t=1.8 \;\rm s$

Distance covered at 1.8 s is,

$x'=\dfrac{5}{2}(1.8)^{2}-9(1.8)\\x'=-8.1 \;\rm m$

So from time interval 0 to 1.8 seconds, the distance covered is 8.1 m.

Distance covered at ( t' = 3 - 1.8 =1.2 s) is,

$x''=\dfrac{5}{2}t^{2}-9t \\x''=\dfrac{5}{2}(1.2)^{2}-9(1.2)\\x''= -7.2 \;\rm m$

Then, total distance is,

$D=x'+x''\\D=8.1 + 7.2\\D=15.3 \;\rm m$

Thus, the distance covered during the given time interval is 15.3 m.

For more details, refer the link:

https://brainly.com/question/11205782?referrer=searchResults