Question

A block starts from rest and slides down an inclined plane with an acceleration of 2m/s^2. How far will it go during the first 2 seconds? What will be its velocity at the end of the 3rd second?

Answer 1

To solve this problem, we can use the equations of motion for objects moving along an inclined plane. The key equation we'll use is:

$s=ut+\frac{1}{2}a{t}^2$

where: s = distance traveled u = initial velocity (0 m/s as the block starts from rest) a = acceleration (2 m/s^2) t = time

To find the distance traveled during the first 2 seconds, we can substitute the given values into the equation:

$s=(0\text{m/s})\cdot (2\text{s})+\frac{1}{2}\cdot (2{\text{m/s}}^2)\cdot (2\text{s}{)}^2$

$s=0+\frac{1}{2}\cdot 2\cdot 4$

$s=0+4$

Therefore, the block will travel a distance of 4 meters during the first 2 seconds.

To find the velocity at the end of the 3rd second, we need to calculate the distance traveled during the first 3 seconds. We'll again use the equation:

s=ut+12at2s = ut + frac{1}{2}at^2<spa