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A particle moves with a constant acceleration `a=2ms^-2` along a straight line. If it moves with an initial velocity of `5ms^-1`, then obtain an expression for its instantaneous velocity. |
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Answer» We know that acceleration is the time rate of change of velocity, i.e. `a=dv//dt` and differentiation is the inverse operation of integration. So by integrating acceleration, we can obtain the expression of velocity. So, `v=intadt=2intdt=2t+c` where c is the constant of integration and its value can be obtained from the initial conditions. At `t=0`, `v=5ms^-1`. Putting these in Eq. (i), we have `5=2xx0+c` `impliesc=5ms^-1` Therefore, `v=2t+5` is the required expression for the instantaneous velocity. |
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