Question

A body moving with constant acceleration travels the distances 84m, 264 m in 6s and 11s respectively. Calculate the initial velocity and acceleration.​

Answer 1

Answer:

Initial velocity = 2 m/sec and Acceleration = 4 m/s²

Explanation:

Given:-

A body moving with constant acceleration travels the distances 84m, 264 m in 6s and 11s respectively.

To Find:-

Initial velocity and acceleration.

Solution:-

♦ By the second equation of motion

$\mapsto \boxed{ \purple{ \bf \: s = ut = \frac{1}{2} a {t}^{2} }}$

♣ For 1st case

$\leadsto \rm \: 84 = u \times 6 + \frac{1}{2} \times (a) \times {6}^{2} \\ \leadsto \rm \: 84 = 6u + 18 a \\ \leadsto \rm \: \color{navy}u + 3a = 14 - - - (i)$

♠ For 2nd case

$\longrightarrow \rm \: 264 = u \times 11 + \frac{1}{2} \times (a) \times {11}^{2} \\ \longrightarrow \rm \: 264 = 11u + \frac{1}{2} \times (a) \times 121 \\ \longrightarrow \rm \: 22u + 121a = 528 \\ \longrightarrow \rm \: \color{aqua}2u + 11a = 48 - - - (ii)$

★ Now, Multiplying (i) by 2 and then subtracting it from (ii)

$: \longmapsto \rm \:(u + 3a = 14) \times 2 \\ : \longmapsto \rm \: \purple{(2u + 6a + 28)}$

Now,

$\rm \: 2u + 11a = 48 \\ \rm 2u + 6a = 28 \\ - \: \: \: \: - \: \: \: \: \: \: \: \: \: - \\ - - - - - - \\ \rm 5a = 20 \\ \color{maroon} \boxed{ \bf \: a = 4}$

Now, Putting the value of a in 1st equation-

→ u + 3a = 14

→ u + (3 × 4) = 14

→ u + 12 = 14

→ u = 14 - 12

→ u = 2

So,

Initial velocity = 2 m/sec and Acceleration = 4 m/s²