An object starts moving with a velocity $6 m / s$ and acceleration $105 m / s^{2}$ after. What time will it acquire the velocity $15 m / s$ ? Find the distance traveled by the object in this time.

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Solution

By using the equation of motion

$v = u + a t$

By substituting the given values in the above equation

$15 = 6 + 105 t$

Then the time is $0.0857 sec$

The distance covered is given as

$S = u t + \frac{1}{2} a t^{2}$

$S = 6 \times 0.0857 + \frac{1}{2} \times 105 \times {(0.0857)}^{2}$

$S = 0.514 + 0.385$

The distance covered is $0.899 m$

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An object starts moving with a velocity 6m/s and acceleration 105m/s2 after. What time will it acquire the velocity 15m/s ? Find the distance traveled by the object in this time.

By using the equation of motion v=u+at By substituting the given values in the above equation 15=6+105t Then the time is0.0857sec The distance covered is given as S=ut+12at2 S=6×0.0857+12×105×(0.0857)2 S=0.514+0.385 The distance covered is0.899m

An object starts moving with a velocity $6 m / s$ and acceleration $105 m / s^{2}$ after. What time will it acquire the velocity $15 m / s$ ? Find the distance traveled by the object in this time.