Question

the initial velocity of a body moving along a straight line is 8u. it has uniform acceleration of 6 m/s square. the distance covered by the body in the 6th second of its motion is 41m. the value of u is
(a) 16m/s
(b) 12m/s
(c) 8m/s
(d) 1m/s

Answer 1

Solution :

⏭ Given:

✏ Initial velocity of body = 8u

✏ Acceleration = 6 $\sf{ms^{-2}}$

✏ Distance covered by body in the 6th second of its motion = 41m

⏭ To Find:

✏ The value of u

⏭ Formula:

✏ Formula of distance covered in $\sf{n^{th}}$ second of its motion is given by

$\star \: \underline{ \boxed{ \bold{ \sf{ \pink{d_{ {n}^{th} } = u + \frac{a}{2} (2n - 1)}}}}} \: \star$

⏭ Terms indication:

✏ d denotes distance covered in $\sf{n^{th}}$ second

✏ u denotes initial velocity

✏ a denotes acceleration

✏ n denotes no. of second

⏭ Calculation:

$\mapsto \sf \: 41 = 8u + \frac{6}{2} (12 - 1) \\ \\ \mapsto \sf \: 41 = 8u + 3(11) \\ \\ \mapsto \sf \: 41 - 33 = 8u \\ \\ \mapsto \sf \: 8u = 8 \\ \\ \mapsto \sf \: u = \frac{ \cancel{8}}{ \cancel{8}} \\ \\ \mapsto \: \boxed{ \boxed{ \large \bold{ \sf{ \purple{u = 1 \: mps}}}}} \: \orange{ \bigstar}$

Answer 2

Answer:

d

Explanation: