An object moves along the x axis according to the equation x = 2.95t^2 - 2.00t +

Question

An object moves along the x axis according to the equation x = 2.95t^2 - 2.00t + 3.00, where x is in meters and t is in seconds. (a) Determine the average speed between t = 3.00 s and t = 4.60 s. (b) Determine the instantaneous speed at t = 3.00s m/s. Determine the instantaneous speed at t = 4.60 s. (c) Determine the average acceleration between t = 3.00 s and t = 4.60 s. (d) Determine the instantaneous acceleration at t = 3.00 s. Determine the instantaneous acceleration at t = 4.60 s. (e) At what time is the object at rest?

Explanation / Answer

Given that -

x(t) = (2.95 t^2 - 2.00t + 3.00) m

So,

v(t) = dx/dt = speed = (2.95*2 t - 2 ) m/s

=> v(t) = (5.9 t - 2 ) m/s

And -

acceleration, a(t) = dv/dt = 5.9

(a) Average speed between t = 3.0 s and t = 4.60 s.
v(aver) = [v(3.0) + v(4.6)]/2 = [(5.9*3.0 - 2 ) + (5.9*4.6 - 2 )]/2
=> v(aver) = 20.42 m/s

(b) Instantaneous speed at t = 3.0 s
v(3.0) = (5.9*3.0 - 2 ) = 15.7 m/s

v(4.6) = (5.9*4.6 - 2) = 25.14 m/s

(c) Average acceleration between t = 3.0 and t = 4.6 s
a = dv/dt = 5.9 m/s^2 = constant

(d) Instantaneous acceleration at t = 3.0 s

= 5.9 m/s^2

And the value will be the same for t = 4.6 s

(e) The object is moving with constant acceleration of value 5.9 m/s^2.

So, it will come at rest at t = infinity seconds.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---