If the distance teavelled by the body is given by 2s=10t+2tsquare.Then the acceleration of the body is
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a= 2(t-1)
dv/dt=2t-2
dv=(2t-2)dt
Inergravity both sides we get,
v= t^ 2-2t
Now,the required magnitude of velocity drives by simply putting the time given (5 seconds) in the above question
Answer: {5^2-(2×5)}=15 m
dv/dt=2t-2
dv=(2t-2)dt
Inergravity both sides we get,
v= t^ 2-2t
Now,the required magnitude of velocity drives by simply putting the time given (5 seconds) in the above question
Answer: {5^2-(2×5)}=15 m
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