Question

7. A body starts moving from rest and moves with a uniform acceleration of 5 cm/^3. In how much time
the velocity of body will become 50 cm/s and how much will it travel in this line?​

Answer 1

time taken = 10 s

distance travelled = 2.5 m

Step-by-step explanation:

Given :

A body starts from rest

Initial velocity = u = 0 cm/s

Acceleration (uniform) = a = 5 cm/s²

Final velocity = v = 50cm/s

To find :

time taken

distance travelled

Solution :

We know that, from the equations of motion,

$\boxed{\begin{array}{cc} \mathrm{v = u + at}\\\text{where}\\\text{v = final velocity}\\\text{u = initial velocity}\\\text{a = acceleration}\\\text{t = time taken}\end{array}}$

So, substituting the values we get,

$\implies \mathrm{50 = (0)+5t}$

$\implies \mathrm{5t = 50}$

$\implies \mathrm{t = \dfrac{50}{5}}$

$\therefore \underline{\boxed{\mathbf{t = 10\ secs}}}$

Thus, time taken to travel in that line is 10 seconds

Now,

We know that, from the equations of motion,

$\boxed{\begin{array}{cc} \mathrm{s = ut +\dfrac{1}{2} at^2}\\\text{where}\\\text{s = displacement}\\\text{u = initial velocity}\\\text{a = acceleration}\\\text{t = time taken}\end{array}}$

So, substituting the values we get,

$\implies \mathrm{s = (0)(10)+\dfrac{1}{2}\times 5 \times (10)^2}$

$\implies \mathrm{s = 25 \times 10}$

$\implies \mathrm{s = 250\ cm}$

$\therefore \underline{\boxed{\mathbf{s = 2.5\ m}}}$

Thus, it travels 2.5m in that line

❖ Extra information

⟡ Equations of motion

$\boxed{\begin{array}{cc}\boxed{\begin{array}{cc}\mathrm{v = u + at}\\ \mathrm{s = ut +\dfrac{1}{2} at^2}\\\mathrm{v^2-u^2 = 2as}\end{array}}\\\text{where}\\\text{v = final velocity}\\\text{s = displacement}\\\text{u = initial velocity}\\\text{a = acceleration}\\\text{t = time taken}\end{array}}$

Hope it helps!!