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प्रश्न
A car starts from rest and moves along the x-axis with constant acceleration 5 ms–2 for 8 seconds. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from the rest?
उत्तर
For constant acceleration of car,
u = 0, a = 5 ms−2, t = 8 s
using s = `"ut" + 1/2 "at"^2`, we get
s1 = `0 + 1/2 xx 5 xx 8^2`
= 160 m
For constant velocity of car,
v = u + at
= 0 + 5 × 8
= 40 ms−1
t = (12 − 8) s = 4 s, a = 0
∴ s2 = vt
= 40 × 4
= 160 m
There fore, total distance covered in 12 s
= s1 + s2
= 160 + 160
= 320 m
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संबंधित प्रश्न
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help of a diagram.
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| Displacement (in m) | 0.05 | 0.20 | 0.45 | 0.80 | 1.25 | 1.80 |
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| Time (s) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| Distance travelled by A (m) | 0 | 20 | 80 | 180 | 240 | 300 | 360 | 420 |
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