Question

10) A body starts from rest with a uniform acceleration of 2ms¯² for 10s, it moves with constant speed for 30s then decelerates by 4ms¯² to zero. what is the distance covered by body?​

Answer 1

Answer: 750 m

Explanation:

Answer 2

$distance \: travelled \: by \: the \: body \: during \: uniform \: acceleration \\ s = ut + \frac{1}{2}a {t}^{2} \\ s = \frac{1}{2} \times 2 \times {10}^{2} \\ s = 100m \\ final \: velocity \: of \: the \: body \: for \: 10seconds \\ v = u + at \\ v = 0 + 2 \times 10 \\ v = 20 \frac{m}{s} \\ distance \: covered \: by \: the \: body \: when \: the \: body \\ moves \: with \: constant \: velocity \\ s = ut + \frac{1}{2}a {t}^{2} \\ s = 20 \times 30 + 0 \\ s = 600m \\ time \: taken \: by \: the \: body \: to \: decelerate \: and \: come \: to \: rest \\ v = u + at \\ 0 = 20 - 4t \\ t = 5sec \\ distance \: travelled \: by \: the \: body \: during \: decleration \\ s = ut + \frac{1}{2}a {t}^{2} \\ s = 20 \times 5 - \frac{1}{2} \times 4 \times {5}^{2} \\ s = 100 - 50 \\ s = 50m \\ total \: distance \: travelled = 100 + 600 + 50 = 750m$

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