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