Question

1. a bird started from rest and accelerated at a constant rate to cover a displacement of 28 m in 11s. what is its final velocity after 11 s?
2. a police car is traveling at 52.0 km/h when a speeding car races past. the police car accelerates at 5.24 m/s², reaching a final velocity of 108 km/h.

a. how long did it take the police car to reach full speed?
b. how far did it travel in this time?

1. use the graph below to answer the following questions

a. when is the object slowing down?
b. find average acceleration from 10 - 55 min?
c. what is the displacement from 0 to 30 min?

Explanation:

3.

Step1: 選用位移公式

已知初速度\(u = 0\)（from rest），位移\(s=28m\)，時間\(t = 11s\)，根據位移公式\(s=ut+\frac{1}{2}at^{2}\)，因為\(u = 0\)，所以\(s=\frac{1}{2}at^{2}\)，可求出加速度\(a\)。
\[a=\frac{2s}{t^{2}}=\frac{2\times28}{11^{2}}=\frac{56}{121}m/s^{2}\]

Step2: 用速度公式求末速度

根據速度公式\(v = u+at\)，\(u = 0\)，\(a=\frac{56}{121}m/s^{2}\)，\(t = 11s\)，則\(v=at\)。
\[v=\frac{56}{121}\times11=\frac{56}{11}\approx5.09m/s\]

Step1: 單位轉換

將初速度\(u = 52.0km/h\)和末速度\(v = 108km/h\)轉換為\(m/s\)。
\(u = 52.0\times\frac{1000}{3600}=\frac{130}{9}m/s\approx14.44m/s\)，\(v = 108\times\frac{1000}{3600}=30m/s\)，加速度\(a = 5.24m/s^{2}\)。

Step2: 用速度公式求時間

根據\(v = u+at\)，可得\(t=\frac{v - u}{a}\)。
\[t=\frac{30-\frac{130}{9}}{5.24}=\frac{\frac{270 - 130}{9}}{5.24}=\frac{\frac{140}{9}}{5.24}=\frac{140}{9\times5.24}\approx2.96s\]

Step1: 選用位移公式

已知\(u=\frac{130}{9}m/s\)，\(a = 5.24m/s^{2}\)，\(t\approx2.96s\)，根據位移公式\(s=ut+\frac{1}{2}at^{2}\)。
\[s=\frac{130}{9}\times2.96+\frac{1}{2}\times5.24\times(2.96)^{2}\]
\[s=\frac{384.8}{9}+ \frac{1}{2}\times5.24\times8.7616\]
\[s\approx42.76+22.99 = 65.75m\]

Answer:

\(\frac{56}{11}m/s\)

4. a.