Question

Suppose that the position vector for a particle is given as a function of time by (t) = x(t)î + y(t)ĵ, with x(t) = at + b and y(t) = ct² + d, where a = 1.40 m/s, b = 1.50 m, c = 0.121 m/s², and d = 1.18 m.

(a) Calculate the average velocity during the time interval from t = 2.10 s to t = 3.90 s.

=

m/s

(b) Determine the velocity at t = 2.10 s.

=

m/s

Determine the speed at t = 2.10 s.
m/s

Answer 1

Part (a)

At t = 2.10 second, the position of the particle will be

$r(2.1) = (at + b) i + (ct^2 + d)j$

$r(2.1) = (1.4 \times 2.1+ 1.5) i + (0.121 \times 2.1^2 + 1.18)j$

$r(2.1) = 4.44 i + 1.71j$

Similarly, t = 3.90 second, the position will be

$r(3.9) = (1.4 \times 3.9+ 1.5) i + (0.121 \times 3.9^2 + 1.18)j$

$r(3.9) = 6.96i + 3.02j$

Hence, the average velocity is given by

$v = \frac{dr}{dt}$

v = { (6.96 - 4.44)i + (3.02 - 1.71) j} ÷ (3.90 - 2.1)

v = (2.52i + 1.31j) ÷ (1.8)

v = 1.4i + 0.72j

In terms of magnitude, it will be

$v = \sqrt{ 1.4^2 + 0.72^2}$

v = 1.57 m/s

Part B)

The instantaneous velocity is also given by

$v = \frac{dr}{dt}$

differentiating w.r.t t

$v = ai + 2ctj$

So, the velocity will be at t = 2.1 is

$v = 1.4i + 0.5j$

The speed will be

$s = \sqrt{ 1.4^2 + 0.5^2}$

s = 1.48 m/s