Question

1D Kinematics Problem: An object moves along the x axis according to the equation

x = 2.80t² − 2.00t + 3.00,

where x is in meters and t is in seconds.

(a) Determine the average speed between t = 2.10 s and t = 3.40 s.


(b) Determine the instantaneous speed at t = 2.10 s.


Determine the instantaneous speed at t = 3.40 s.


(c) Determine the average acceleration between t = 2.10 s and t = 3.40 s.


(d) Determine the instantaneous acceleration at t = 2.10 s.

Determine the instantaneous acceleration at t = 3.40 s.


(e) At what time is the object at rest?

Answer 1

a)

position at any time is given as

x = 2.80 t² − 2.00 t + 3.00

at t = 2.10 s, position is given as

x_i = 2.80 (2.10)² − 2.00 (2.10) + 3.00 = 11.15 m

at t = 3.40 s, position is given as

x_f = 2.80 (3.40)² − 2.00 (3.40) + 3.00 = 28.57 m

distance travelled = d = x_f - x_i = 28.57 - 11.15 = 17.42 m

t = time travelled = 3.40 - 2.10 = 1.3 sec

average speed is given as

V_avg = d/t = 17.42/1.3 = 13.4 m/s

b)

x = 2.80 t² − 2.00 t + 3.00

taking derivative both side relative to "t"

dx/dt = 5.6 t - 2

instanteneous speed is given as

v = dx/dt = 5.6 t - 2

at t = 2.10 , instanteneous speed is given as

v_i = 5.6 (2.10) - 2 = 9.76 m/s

at t = 3.40 , instanteneous speed is given as

v_f = 5.6 (3.40) - 2 = 17.04 m/s

c)

average acceleration is given as

a_avg = change in velocity/ time interval = (v_f - v_i)/t = (17.04 - 9.76)/1.3 = 5.6 m/s²

d)

instantenous acceleration is given as

a = d²x/dt² = dv/dt = d/dt (5.6 t - 2) = 5.6 m/s²

at t = 2.10 s

a = 5.6 m/s²

at t = 3.40 s

a = 5.6 m/s²

e)

object is at rest when instanteous speed is zero

hence v = 0

5.6 t - 2 = 0

t = 0.36 sec