The velocity graph of a particle moving along the x-axis is shown. The particle has zero velocity at t=0.00s and reaches a maximum velocity, vmax, after a total elapsed time, t total. If the initial position of the particle is x0 =6.22m, the maximum velocity of the particle is vmax=47.6m/s, and the total elapsed time is t total=17.6s, what is the particle's position at t=11.7s?
At t=11.7s, what is the particle's velocity?
At t=11.7s, what is the particle's acceleration?
the graph is a linear line increasing from origin
The area under the graph from t=0 to the time asked for is distance. Since it is a straight line, use your area formulas to figure it out. Distance is the area from y=0 t the graph, from t0 to time t.
let me try this again area formula is 1/2 x b x h
1/2 x 2.705 x 6.22 m/s = 8.4125 m
is this correct? so after 11.7 seconds the particle position will be 8.4125 m?
To find the particle's position at t = 11.7s, we need to integrate the velocity function over the interval [0, 11.7]. Since the velocity graph is a linear line increasing from the origin, the velocity function can be represented by the equation: v(t) = mt, where "m" is the slope of the line.
To find "m", we can use the maximum velocity and total elapsed time given in the question. From the graph, we know that the maximum velocity, vmax, occurs at t = t total. Therefore, at t = 17.6s, v(17.6) = vmax = 47.6 m/s.
Using this information, we can calculate the slope "m" as follows:
m = vmax / t total
= 47.6 m/s / 17.6 s
= <<47.6/17.6=2.704>>2.70 m/s^2
Now that we know the slope "m", we can find the position at t = 11.7s:
x(t) = x0 + ∫(0 to t) v(t') dt'
= x0 + ∫(0 to 11.7) (mt') dt'
Evaluating the integral, we get:
x(11.7) = x0 + (m/2) * (t')^2 |from 0 to 11.7
= x0 + (2.70/2) * (11.7)^2
Using the given initial position x0 = 6.22m, we can calculate the particle's position at t = 11.7s:
x(11.7) = 6.22 + (2.70/2) * (11.7)^2
= 6.22 + (2.70/2) * 136.89
= 6.22 + 183.06075
= 189.28075 m
Therefore, the particle's position at t = 11.7s is approximately 189.28 m.
To find the particle's velocity at t = 11.7s, we can plug the value of t into the velocity function:
v(11.7) = m * 11.7
= 2.70 m/s^2 * 11.7 s
≈ 31.59 m/s
Therefore, the particle's velocity at t = 11.7s is approximately 31.59 m/s.
To find the particle's acceleration at t = 11.7s, we need to differentiate the velocity function:
a(t) = d(v(t))/dt
= d(mt)/dt
= m
Since "m" represents the slope of the velocity graph, it remains constant over time. Therefore, the particle's acceleration at t = 11.7s is the same as the slope "m" calculated earlier:
a(11.7) = m
= 2.70 m/s^2
Therefore, the particle's acceleration at t = 11.7s is 2.70 m/s^2.
a=47.6-0/17.6
a=2.705 seconds