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29 January, 00:54

a body initially at rest, starts moving with a constant acceleration of 2ms-2. calculate the velocity acquired and the distance travelled in 5s

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  1. 29 January, 04:42
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    The velocity acquired in 5 s is 10 m/s and the distance travelled in 5 s is 25 m.

    Explanation:

    As acceleration is the measure of change in velocity with respect to time, if the acceleration and time is known, we can easily determine the velocity using the first equation of motion.

    v = u + at

    Here v is the final velocity which we have to determine, u is the initial velocity which is given as zero and acceleration is given as 2 ms⁻².

    So, velocity of the body at 5 s will be v = 0 + (2 * 5 s), v = 10 m/s.

    As the given problem stated that motion is occurring at constant acceleration, then the distance travelled by the body can be obtained from second equation of motion.

    s = ut + 0.5 at²

    So, here the displacement s will be equal to the distance as the starting position is considered as zero. And initial velocity u is also zero, a is given as 2 ms⁻² and the time is 5s.

    Then, s = (0*5) + (0.5*2*5*5) = 25 m

    So, the velocity acquired in 5 s is 10 m/s and the distance travelled in 5 s is 25 m.
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