The energy required to accelerated a car from 10m/s to 20m/s is how many times the energy required to accelerate the car from rest to 10m/s
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Answered by
24
Energy required to accelerate from 10m/s to 20m/s is KE1 and KE2 is the energy required to accelerate from rest to 10m/s
KE1=
![\frac{1}{2} m {v}^{2} - \frac{1}{2} m {u}^{2} \\ \frac{1}{2} m( {20}^{2} - {10}^{2} ) \\ 150m \frac{1}{2} m {v}^{2} - \frac{1}{2} m {u}^{2} \\ \frac{1}{2} m( {20}^{2} - {10}^{2} ) \\ 150m](https://tex.z-dn.net/?f=+%5Cfrac%7B1%7D%7B2%7D+m+%7Bv%7D%5E%7B2%7D++-++%5Cfrac%7B1%7D%7B2%7D+m+%7Bu%7D%5E%7B2%7D++%5C%5C++%5Cfrac%7B1%7D%7B2%7D+m%28++%7B20%7D%5E%7B2%7D++-++%7B10%7D%5E%7B2%7D+%29+%5C%5C+150m)
KE2=
![\frac{1}{2} m {v}^{2} - \frac{1}{2} m {u}^{2} \\ \frac{1}{2} m( {v}^{2} - {u}^{2} ) \\ \frac{1}{2} m( {10}^{2} - 0) \\ \frac{1}{2} m100 = 50m \frac{1}{2} m {v}^{2} - \frac{1}{2} m {u}^{2} \\ \frac{1}{2} m( {v}^{2} - {u}^{2} ) \\ \frac{1}{2} m( {10}^{2} - 0) \\ \frac{1}{2} m100 = 50m](https://tex.z-dn.net/?f=+%5Cfrac%7B1%7D%7B2%7D+m+%7Bv%7D%5E%7B2%7D++-++%5Cfrac%7B1%7D%7B2%7D+m+%7Bu%7D%5E%7B2%7D++%5C%5C++%5Cfrac%7B1%7D%7B2%7D+m%28+%7Bv%7D%5E%7B2%7D++-++%7Bu%7D%5E%7B2%7D+%29+%5C%5C++%5Cfrac%7B1%7D%7B2%7D+m%28+%7B10%7D%5E%7B2%7D++-+0%29+%5C%5C++%5Cfrac%7B1%7D%7B2%7D+m100+%3D+50m)
KE1=150m
KE2=50m
so KE1=3KE2.
hence answer is 1/3 times energy required to make body accelerate from rest to 10m/s than from 10m/s to 20m/s.
KE1=
KE2=
KE1=150m
KE2=50m
so KE1=3KE2.
hence answer is 1/3 times energy required to make body accelerate from rest to 10m/s than from 10m/s to 20m/s.
Answered by
17
Answer:
3 times
answer is given in the attachment
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