Obtain a relation for the distance travelled | vedclass

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Question

(C) The distance travelled by an object in time $t$ with uniform acceleration $a$ and initial velocity $u$ is given by the equation of motion: $s = ut

Vedclass · Accepted Answer

$(u+\frac{9}{2}a) \, m$

Vedclass · Answer

(C) The distance travelled by an object in time $t$ with uniform acceleration $a$ and initial velocity $u$ is given by the equation of motion: $s = ut + \frac{1}{2}at^2$.First,calculate the total distance travelled in $5 \, s$:$s_5 = u(5) + \frac{1}{2}a(5)^2 = 5u + \frac{25}{2}a$ ........... $(i)$Next,calculate the total distance travelled in $4 \, s$:$s_4 = u(4) + \frac{1}{2}a(4)^2 = 4u + \frac{16}{2}a = 4u + 8a$ ........... $(ii)$The distance travelled in the interval between the $4^{th}$ and $5^{th}$ second is the difference between the distance at $5 \, s$ and $4 \, s$:Distance $= s_5 - s_4 = (5u + \frac{25}{2}a) - (4u + 8a)$Distance $= (5u - 4u) + (12.5a - 8a) = u + 4.5a$Distance $= (u + \frac{9}{2}a) \, m$.

Obtain a relation for the distance travelled by an object moving with a uniform acceleration in the interval between the $4^{th}$ and $5^{th}$ seconds.

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