Answer :
To solve this problem, we break it down into three phases - acceleration, constant speed, and deceleration:
### Phase 1: Acceleration
1. Initial velocity ([tex]\(u_1\)[/tex]): 0 m/s
2. Final velocity ([tex]\(v\)[/tex]): 10 m/s
3. Time ([tex]\(t_1\)[/tex]): [tex]\(2T\)[/tex] seconds
4. Acceleration ([tex]\(a\_accel\)[/tex]): Use the formula [tex]\( v = u_1 + a\_accel \cdot t_1 \)[/tex]
[tex]\[ 10 = 0 + a\_accel \cdot 2T \implies a\_accel = \frac{10}{2T} = \frac{5}{T} \, \text{m/s}^2 \][/tex]
5. Distance covered during acceleration ([tex]\(s_1\)[/tex]): Use the formula [tex]\( s = u_1 \cdot t_1 + \frac{1}{2} a\_accel \cdot t_1^2 \)[/tex]
[tex]\[ s_1 = 0 \cdot 2T + \frac{1}{2} \cdot \frac{5}{T} \cdot (2T)^2 = \frac{1}{2} \cdot \frac{5}{T} \cdot 4T^2 = 10T \, \text{meters} \][/tex]
### Phase 2: Constant Speed
1. Speed ([tex]\(v\)[/tex]): 10 m/s
2. Time ([tex]\(t_2\)[/tex]): 60 seconds
3. Distance covered during constant speed ([tex]\(s_2\)[/tex]): Use the formula [tex]\( s = v \cdot t_2 \)[/tex]
[tex]\[ s_2 = 10 \cdot 60 = 600 \, \text{meters} \][/tex]
### Phase 3: Deceleration
1. Initial velocity ([tex]\(u_3\)[/tex]): 10 m/s
2. Final velocity ([tex]\(v\_final\)[/tex]) [tex]\( = 0\)[/tex] m/s
3. Time ([tex]\(t_3\)[/tex]): [tex]\(8T\)[/tex] seconds
4. Deceleration ([tex]\(a_\text{decel}\)[/tex]): Use the formula [tex]\( v\_final = u_3 - a_\text{decel} \cdot t_3 \)[/tex]
[tex]\[ 0 = 10 - a\_decel \cdot 8T \implies a\_decel = \frac{10}{8T} = \frac{5}{4T} \, \text{m/s}^2 \][/tex]
5. Distance covered during deceleration ([tex]\(s_3\)[/tex]): Use the formula [tex]\( s = u_3 \cdot t_3 - \frac{1}{2} a_\text{decel} \cdot t_3^2 \)[/tex]
[tex]\[ s_3 = 10 \cdot 8T - \frac{1}{2} \cdot \frac{5}{4T} \cdot (8T)^2 = 80T - \frac{1}{2} \cdot \frac{5}{4T} \cdot 64T^2 = 80T - 40T = 40T \, \text{meters} \][/tex]
### Total Distance and Solving for [tex]\( T \)[/tex]
The total distance travelled by the car is the sum of the distances covered in all three phases:
[tex]\[ s_\text{total} = s_1 + s_2 + s_3 = 10T + 600 + 40T = 50T + 600 \, \text{meters} \][/tex]
Given that the total distance is 800 meters, we set up the equation:
[tex]\[ 50T + 600 = 800 \][/tex]
Solving for [tex]\( T \)[/tex]:
[tex]\[ 50T = 800 - 600 \][/tex]
[tex]\[ 50T = 200 \][/tex]
[tex]\[ T = \frac{200}{50} = 4 \, \text{seconds} \][/tex]
Thus, the value of [tex]\( T \)[/tex] is [tex]\( \boxed{4} \)[/tex] seconds.
### Phase 1: Acceleration
1. Initial velocity ([tex]\(u_1\)[/tex]): 0 m/s
2. Final velocity ([tex]\(v\)[/tex]): 10 m/s
3. Time ([tex]\(t_1\)[/tex]): [tex]\(2T\)[/tex] seconds
4. Acceleration ([tex]\(a\_accel\)[/tex]): Use the formula [tex]\( v = u_1 + a\_accel \cdot t_1 \)[/tex]
[tex]\[ 10 = 0 + a\_accel \cdot 2T \implies a\_accel = \frac{10}{2T} = \frac{5}{T} \, \text{m/s}^2 \][/tex]
5. Distance covered during acceleration ([tex]\(s_1\)[/tex]): Use the formula [tex]\( s = u_1 \cdot t_1 + \frac{1}{2} a\_accel \cdot t_1^2 \)[/tex]
[tex]\[ s_1 = 0 \cdot 2T + \frac{1}{2} \cdot \frac{5}{T} \cdot (2T)^2 = \frac{1}{2} \cdot \frac{5}{T} \cdot 4T^2 = 10T \, \text{meters} \][/tex]
### Phase 2: Constant Speed
1. Speed ([tex]\(v\)[/tex]): 10 m/s
2. Time ([tex]\(t_2\)[/tex]): 60 seconds
3. Distance covered during constant speed ([tex]\(s_2\)[/tex]): Use the formula [tex]\( s = v \cdot t_2 \)[/tex]
[tex]\[ s_2 = 10 \cdot 60 = 600 \, \text{meters} \][/tex]
### Phase 3: Deceleration
1. Initial velocity ([tex]\(u_3\)[/tex]): 10 m/s
2. Final velocity ([tex]\(v\_final\)[/tex]) [tex]\( = 0\)[/tex] m/s
3. Time ([tex]\(t_3\)[/tex]): [tex]\(8T\)[/tex] seconds
4. Deceleration ([tex]\(a_\text{decel}\)[/tex]): Use the formula [tex]\( v\_final = u_3 - a_\text{decel} \cdot t_3 \)[/tex]
[tex]\[ 0 = 10 - a\_decel \cdot 8T \implies a\_decel = \frac{10}{8T} = \frac{5}{4T} \, \text{m/s}^2 \][/tex]
5. Distance covered during deceleration ([tex]\(s_3\)[/tex]): Use the formula [tex]\( s = u_3 \cdot t_3 - \frac{1}{2} a_\text{decel} \cdot t_3^2 \)[/tex]
[tex]\[ s_3 = 10 \cdot 8T - \frac{1}{2} \cdot \frac{5}{4T} \cdot (8T)^2 = 80T - \frac{1}{2} \cdot \frac{5}{4T} \cdot 64T^2 = 80T - 40T = 40T \, \text{meters} \][/tex]
### Total Distance and Solving for [tex]\( T \)[/tex]
The total distance travelled by the car is the sum of the distances covered in all three phases:
[tex]\[ s_\text{total} = s_1 + s_2 + s_3 = 10T + 600 + 40T = 50T + 600 \, \text{meters} \][/tex]
Given that the total distance is 800 meters, we set up the equation:
[tex]\[ 50T + 600 = 800 \][/tex]
Solving for [tex]\( T \)[/tex]:
[tex]\[ 50T = 800 - 600 \][/tex]
[tex]\[ 50T = 200 \][/tex]
[tex]\[ T = \frac{200}{50} = 4 \, \text{seconds} \][/tex]
Thus, the value of [tex]\( T \)[/tex] is [tex]\( \boxed{4} \)[/tex] seconds.
