Question: A 747 jetliner lands and begins to slow to a stop as it moves along the runway. If its mass is $3.30\times10^{5}$ kg its speed is $27.0$ m/s, and the net braking force is $4.3\times10^{5}$ N, (a) what is its speed $7.50$ s later and (b) how far has it travelled in this time? (a) Speed in m/s, to nearest hundredth.
1. **State the problem:**
We have a jetliner with mass $m=3.30\times10^{5}$ kg moving at initial speed $v_0=27.0$ m/s. A net braking force $F=4.3\times10^{5}$ N slows it down. We want to find its speed after $t=7.50$ s.
2. **Relevant formula:**
Newton's second law: $$F=ma$$ where $a$ is acceleration.
Since the force is braking, acceleration will be negative.
3. **Calculate acceleration:**
$$a=\frac{F}{m}=\frac{-4.3\times10^{5}}{3.30\times10^{5}}=-1.30303...\text{ m/s}^2$$
4. **Calculate speed after $7.50$ s:**
Using the formula for velocity under constant acceleration:
$$v=v_0+at$$
Substitute values:
$$v=27.0 + (-1.30303...) \times 7.50$$
$$v=27.0 - 9.7727 = 17.2273\text{ m/s}$$
5. **Round to nearest hundredth:**
$$v=17.23\text{ m/s}$$
**Final answer:** The speed after $7.50$ s is $\boxed{17.23}$ m/s.