A car starts from rest and moves with constant acceleration of 10m/s2 for a time of 5 seconds, after which it moves with constant velocity achieved for another time of 5 seconds after which it retards with retardation of 5m/s2 and comes to rest. Calculate its maximum speed during the journey and total distance travelled.
Answers
Answer:
Maximum speed = 50 m/s
Total distance = 1000 m or 1 km
Explanation:
The maximum speed is attained just before the car starts to exhibit retardation. According to the question, the speed attained before retardation is the constant velocity. Hence the maximum speed of the car is the constant velocity mentioned.
Constant velocity is obtained after 5 seconds of constant acceleration. Using first equation of motion we get:
→ v = u + at
→ v = 0 + 10 ( 5 )
→ v = 50 m/s
Hence the value of the constant velocity is 50 m/s
Hence the maximum speed of the car is 50 m/s.
Distances traveled by car:
- Distance traveled with constant acceleration - (1)
- Distance traveled with constant velocity - (2)
- Distance traveled with retardation - (3)
Hence Total Distance = (1) + (2) + (3)
Distance travelled with constant acceleration:
→ s = ut + 0.5 at²
→ s = 0(5) + 0.5 ( 10 ) ( 5² )
→ s = 5 × 25 = 125 m ... (i)
Distance traveled with constant velocity:
→ d = v × t
→ d = 50 × 5
→ d = 250 m ...(ii)
Distance travelled with uniform retardation:
→ s = 50t + 0.5 ( -5 ) ( t² ) ...(iii)
To calculate t, we use first equation of motion.
→ v = u + at
→ 0 = 50 - 5t
→ 5t = 50
→ t = 10 seconds
Substituting back in (iii) we get:
→ s = 50 ( 10 ) - 0.5 ( 5 ) ( 100 )
→ s = 500 - 250
→ s = 250 m ...(iv)
Adding all the distances ( i + ii + iv ) we get:
→ Total distance = 250 + 500 + 250
→ Total Distance = 1000 m = 1 km
Hence the total distance traveled by the car in its entire journey is 1 km.
Answer:
Maximum speed = 50 m/s
Total distance = 1000 m or 1 km
Explanation:
The maximum speed is attained just before the car starts to exhibit retardation. According to the question, the speed attained before retardation is the constant velocity. Hence the maximum speed of the car is the constant velocity mentioned.
Constant velocity is obtained after 5 seconds of constant acceleration.
Using first equation of motion we get:
→ v = u + at
→ v = 0 + 10 ( 5 )
→ v = 50 m/s
Hence the value of the constant velocity is 50 m/s.
Hence the maximum speed of the car is 50 m/s.
Distances traveled by car:
1. Distance traveled with constant acceleration.
2. Distance traveled with constant velocity.
3. Distance traveled with retardation.
So, Total Distance = (1) + (2) + (3)
Distance traveled with constant acceleration:
→ s = ut + 0.5 at²
→ s = 0(5) + 0.5 ( 10 ) ( 5² )
→ s = 5 × 25 = 125 m ... 1
Distance traveled with constant velocity:
→ d = v × t
→ d = 50 × 5
→ d = 250 m ... 2
Distance traveled with uniform retardation:
→ s = 50t + 0.5 ( -5 ) ( t² ) ... 3
To calculate t, we use first equation of motion.
→ v = u + at
→ 0 = 50 - 5t
→ 5t = 50
→ t = 10 seconds
Substituting back in (3) we get:
→ s = 50 ( 10 ) - 0.5 ( 5 ) ( 100 )
→ s = 500 - 250
→ s = 250 m ... 4
Adding all the distances (1 + 2 + 4) we get:
→ Total distance = 250 + 500 + 250
→ Total Distance = 1000 m = 1 km
Hence the total distance traveled by the car in its entire journey is 1 km.