Question

An object moves along the x axis according to the equation

x = 3.65t² − 2.00t + 3.00,

where x is in meters and t is in seconds.

(a) Determine the average speed between t = 1.50 s and t = 3.50 s.
m/s

(b) Determine the instantaneous speed at t = 1.50 s.
m/s

Determine the instantaneous speed at t = 3.50 s.
m/s

(c) Determine the average acceleration between t = 1.50 s and t = 3.50 s.
m/s²

(d) Determine the instantaneous acceleration at t = 1.50 s.
m/s²

Determine the instantaneous acceleration at t = 3.50 s.
m/s²

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

Answer 1

(a) The expression is -

x(t) = 3.65*t^2 − 2.00*t + 3.00

so, x(1.5) = 3.65*1.5^2 - 2.0*1.5 + 3.0 = 8.21 m

x(3.5) = 3.65*3.5^2 - 2.0*3.5 + 3.0 = 40.72 m

So, the average speed during this time interval = (40.72 - 8.21) / (3.5 - 1.5) = 32.51 / 2 = 16.26 m/s

(b) v(t) = d(x) / dt = 3.65x2xt - 2 = 7.3*t - 2

v(1.5) = 7.3*1.5 - 2 = 8.95 m/s

v(3.5) = 7.3*3.5 - 2 = 23.55 m/s

(c) acceleration, a(t) = dv/dt = 7.3 m/s^2

This is constant

so average acceleration shall also be constant.

so your answer = 7.3 m/s^2

(d) average acceleration = 7.3 m/s^2 (for both the parts)

(e) when the object at rest -

v(t) = 0

=> 7.3*t - 2 =

=> t = 2 / 7.3 = 0.27 second.