Question

An object has a position given by r⃗ = [2.0 m + (5.00 m/s)t] i^ + [3.0 m − (3.00 m/s2)t2] j^ , where quantities are in SI units. What is the speed of the object at time t = 2.00 s?

Answer 1

Answer:

Speed = 13.0m/s

Step-by-step explanation:

Given that velocity, v = (ai ± bj)m/s.

Because speed is scalar and has only magnitude, we can obtain it from the velocity by calculating only the magnitude given as:

$Speed = (\sqrt{a^{2} + b^{2} } )m/s$      ...Equation 1

From the question, we were given the position vector, r

The velocity can be obtained by differentiating r with respect to time, t

We have that differentiation of a constant is 0.

$v = \frac{dr}{dt} = \frac{d}{dt}[(2.0+5.00t)i + (3.0 - 3.00t^{2})j] \\\\v = 0 + 5.00i + 0 - (6.00t)j$

At time t = 2.00s,

$v = 5.00i - (6.00*2.00)j = 5.00i - 12.00j$

To obtain the speed, we use equation 1

$Speed = \sqrt{(5.00)^{2} + (12.00)^{2} } = \sqrt{25 + 144} \\ \\Speed = \sqrt{169} = 13.00m/s$

Therefore, the speed of the object at time t = 2.00s is 13.00m/s