24. A car is moving at a speed of 40 km/h. 2 seconds after its speed
becomes 60 km. Calculate the acceleration of the car and the distance
travelled.
eed of 54 km/h in 10 s. What will
at
Answers
Answer:
Acceleration = 2.77 m/s²
Distance = 27.77 metres
Explanation:
Given :
- Initial velocity of the car = 40 km/h = u
- Final velocity of the car = 60 km/h = v
- Time taken = 2 seconds
To find :
- Acceleration and the Distance of the car
40 km/h = 40×5/18 = 100/9 m/s
60 km/h = 60×5/18 = 150/9 m/s
Acceleration = (v-u) /t
Acceleration = (150/9 - 100/9)/2
Acceleration = (50/9)/2
Acceleration = 50/9 ×1/2
Acceleration = 25/9 m/s²
Acceleration = 2.777 m/s²
Using the third equation of motion :
V²-u²=2as
(150/9)²-(100/9)²=2×25/9×s
22500/81 - 10000/81 = 50/9 s
12500/81 = 50/9 s
S = 12500/81×9/50
S = 250/9
S = 27.77 metres
Answer:
Explanation:
from the first equation of motion
(note : convert km per hour into metre per second)
v = u + at
16.66 = 11.11 + a * 2
16.66 - 11.11 = 2a
5.55 = 2a
5.55 / 2 = a
Acceleration = 2.77 metre / second ^2
From second equation of motion
s = u * t + 1 / 2 + at^2
s = 11.11 * 2 + 1 / 2 + 2.77 * 2 * 2
s = 22.22 + 5.54
s = 27.76
Distance travelled is = 27.76
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