Question

A body moving with uniform acceleration travels 24 m in the 6th second and 44 m in the 11th second.Find the

(a) Acceleration .

(b) Initial velocity of the body .

Answer 1

Step 1:

Given that

A body is moving with uniform acceleration.

Distance travelled in 6th second is 24 m.

Distance travelled in 11th second is 44 m.

Step 2: Formulas.

Distance $S_{}=ut+\frac{1}{2}\times a\times t^2.$

u = initial velocity

a = acceleration

t = time

Distance travelled in the nth second =$S_n=U+\frac{1}{2}a(2n-1).$

Step 3: Calculation.

(a) Acceleration .

For the 6th second =$24=u+\frac{1}{2}a(2\times 6-1)$.

$24=u+\frac{11}{2}a.\to eqn1$

For the 11th second =$44=u+\frac{1}{2}a(2\times 11-1).$

$44=u+\frac{1}{2}a\left(21\right).$

$44=u+\frac{21}{2}a.\to eqn2.$

Subtracting eqn 1 from eqn 2.

$44-24=\frac{21}{2}a-\frac{11}{2}a.$

$20=\frac{10}{2}a.$

$a=\frac{20\times 2}{10}.$

$a=4\frac{m}{s^2}.$

Step 4: (b) Initial velocity of the body.

Substituting the value of ‘a’ in eqn 2.

$44=u+\frac{21}{2}a.\to eqn2.$

$$\begin{gathered} 44\frac{m}{s}=u+\frac{21}{2}\times 4. \\ 44\frac{m}{s}=u+42\frac{m}{s}. \\ u=44\frac{m}{s}+42\frac{m}{s}. \\ u=2\frac{m}{s}. \end{gathered}$$

Hence,

(a) Acceleration .$a=4\frac{m}{s^2}.$

(b) Initial velocity of the body $u=2\frac{m}{s}.$.