Answer :
Certainly, let's solve this step-by-step:
### 1. Calculate the Mass of the Car
Given:
- Momentum ([tex]\(p\)[/tex]) = 24,000 kg·m/s
- Time after which the momentum is given ([tex]\(t_{\text{momentum}}\)[/tex]) = 3 s
First, let's denote:
- Mass of the car as [tex]\(m\)[/tex]
- Velocity of the car as [tex]\(v\)[/tex]
From the equation of momentum:
[tex]\[ p = m \cdot v \][/tex]
We need to determine the velocity first. Since we are not given the initial velocity directly, we assume that the car started from rest (initial velocity [tex]\( = 0\)[/tex]).
Assuming constant acceleration ([tex]\(a\)[/tex]) over the 3 seconds, we can use the kinematic equation to find velocity:
[tex]\[ v = a \cdot t_{\text{momentum}} \][/tex]
Now, using the formula for momentum:
[tex]\[ p = m \cdot v \][/tex]
Substituting [tex]\(v\)[/tex]:
[tex]\[ p = m \cdot (a \cdot t_{\text{momentum}}) \][/tex]
Given [tex]\(p = 24,000\)[/tex] kg·m/s and [tex]\(t_{\text{momentum}} = 3\)[/tex] s:
[tex]\[ 24,000 = m \cdot (a \cdot 3) \][/tex]
Solving for mass [tex]\(m\)[/tex]:
[tex]\[ m = \frac{24,000}{3a} \][/tex]
[tex]\[ m = \frac{8,000}{a} \][/tex]
### 2. Determine the Distance Travelled by the Car During the First 5 Seconds
To find the distance ([tex]\(d\)[/tex]) travelled during the first 5 seconds ([tex]\(t_{\text{distance}} = 5\)[/tex] s), we use the kinematic equation for distance under constant acceleration:
[tex]\[ d = v_{\text{initial}} \cdot t_{\text{distance}} + \frac{1}{2} \cdot a \cdot t_{\text{distance}}^2 \][/tex]
Since the car started from rest ([tex]\(v_{\text{initial}} = 0\)[/tex]):
[tex]\[ d = 0 \cdot 5 + \frac{1}{2} \cdot a \cdot 5^2 \][/tex]
[tex]\[ d = \frac{1}{2} \cdot a \cdot 25 \][/tex]
[tex]\[ d = 12.5 \cdot a \][/tex]
### Summary of Results
1. Mass of the car ([tex]\(m\)[/tex]):
[tex]\[ m = \frac{8,000}{a} \, \text{kg} \][/tex]
2. Distance travelled ([tex]\(d\)[/tex]):
[tex]\[ d = 12.5 \cdot a \, \text{m} \][/tex]
To get numerical values, we need the value of acceleration [tex]\(a\)[/tex]. If [tex]\(a\)[/tex] is provided or assumed, we can substitute it into the above equations to get the mass and distance.
### 1. Calculate the Mass of the Car
Given:
- Momentum ([tex]\(p\)[/tex]) = 24,000 kg·m/s
- Time after which the momentum is given ([tex]\(t_{\text{momentum}}\)[/tex]) = 3 s
First, let's denote:
- Mass of the car as [tex]\(m\)[/tex]
- Velocity of the car as [tex]\(v\)[/tex]
From the equation of momentum:
[tex]\[ p = m \cdot v \][/tex]
We need to determine the velocity first. Since we are not given the initial velocity directly, we assume that the car started from rest (initial velocity [tex]\( = 0\)[/tex]).
Assuming constant acceleration ([tex]\(a\)[/tex]) over the 3 seconds, we can use the kinematic equation to find velocity:
[tex]\[ v = a \cdot t_{\text{momentum}} \][/tex]
Now, using the formula for momentum:
[tex]\[ p = m \cdot v \][/tex]
Substituting [tex]\(v\)[/tex]:
[tex]\[ p = m \cdot (a \cdot t_{\text{momentum}}) \][/tex]
Given [tex]\(p = 24,000\)[/tex] kg·m/s and [tex]\(t_{\text{momentum}} = 3\)[/tex] s:
[tex]\[ 24,000 = m \cdot (a \cdot 3) \][/tex]
Solving for mass [tex]\(m\)[/tex]:
[tex]\[ m = \frac{24,000}{3a} \][/tex]
[tex]\[ m = \frac{8,000}{a} \][/tex]
### 2. Determine the Distance Travelled by the Car During the First 5 Seconds
To find the distance ([tex]\(d\)[/tex]) travelled during the first 5 seconds ([tex]\(t_{\text{distance}} = 5\)[/tex] s), we use the kinematic equation for distance under constant acceleration:
[tex]\[ d = v_{\text{initial}} \cdot t_{\text{distance}} + \frac{1}{2} \cdot a \cdot t_{\text{distance}}^2 \][/tex]
Since the car started from rest ([tex]\(v_{\text{initial}} = 0\)[/tex]):
[tex]\[ d = 0 \cdot 5 + \frac{1}{2} \cdot a \cdot 5^2 \][/tex]
[tex]\[ d = \frac{1}{2} \cdot a \cdot 25 \][/tex]
[tex]\[ d = 12.5 \cdot a \][/tex]
### Summary of Results
1. Mass of the car ([tex]\(m\)[/tex]):
[tex]\[ m = \frac{8,000}{a} \, \text{kg} \][/tex]
2. Distance travelled ([tex]\(d\)[/tex]):
[tex]\[ d = 12.5 \cdot a \, \text{m} \][/tex]
To get numerical values, we need the value of acceleration [tex]\(a\)[/tex]. If [tex]\(a\)[/tex] is provided or assumed, we can substitute it into the above equations to get the mass and distance.
