
A body starts slipping down an incline and moves half meter in half second. How long will it take to move the next half meter?
Answer
240k+ views
Hint: Use the formula of the displacement in the equation of motion and find the acceleration of the body. Substitute this in the other equation of motion to find the final velocity. Keep that as the initial velocity of the body and substitute this in the first equation of motion to find the value of the time taken.
Formula used:
The equations of motion is given by
$ s = ut + \dfrac{1}{2}a{t^2} \\
v = u + at \\
$
Where $s$ is the distance covered, $u$ is the initial velocity, $a$ is the acceleration of the body, $v$ is the final velocity of the body and $t$ is the time taken for the movement.
Complete step by step solution:
It is given that the
Distance moved, $d = 0.5\,m$
Time taken for travelling the distance of half meter, $t = 0.5\,s$
Using the equation of the motion,
$s = ut + \dfrac{1}{2}a{t^2}$
Substituting the known parameters in the above equation, we get
$\Rightarrow 0.5 = 0t + \dfrac{1}{2}a{\left( {0.5} \right)^2}$
By performing various arithmetic operations, we get
$\Rightarrow a = 4\,m{s^{ - 2}}$
Using the other equation of motion,
$v = u + at$
Substituting the initial velocity, acceleration and the time taken in it
$\Rightarrow v = 0 + 4 \times 0.5 = 2\,m{s^{ - 1}}$
Hence the final velocity of the body is obtained as $2\,m{s^{ - 1}}$ . For the next $0.5\,m$ , the previous final velocity becomes the initial velocity but the acceleration remains the same.
$s = ut + \dfrac{1}{2}a{t^2}$
$\Rightarrow s = 2t + \dfrac{1}{2}4{t^2}$
By simplifying the above equation, we get
$\Rightarrow 2t + 2{t^2} - 0.5 = 0$
By solving the above equation, we get
$\Rightarrow t = 0.207s$
Hence the time taken to cover the second $0.5\,m$ is $0.21\,s$.
Note: The velocity of the body that slides on the inclined surface mainly depends on the friction of the surface, mass of the body and the slope of the inclination of the surface. If the surface is rough and so the friction is more and hence the velocity of the body reduces. If it has a smooth slippery surface, it has a greater velocity.
Formula used:
The equations of motion is given by
$ s = ut + \dfrac{1}{2}a{t^2} \\
v = u + at \\
$
Where $s$ is the distance covered, $u$ is the initial velocity, $a$ is the acceleration of the body, $v$ is the final velocity of the body and $t$ is the time taken for the movement.
Complete step by step solution:
It is given that the
Distance moved, $d = 0.5\,m$
Time taken for travelling the distance of half meter, $t = 0.5\,s$
Using the equation of the motion,
$s = ut + \dfrac{1}{2}a{t^2}$
Substituting the known parameters in the above equation, we get
$\Rightarrow 0.5 = 0t + \dfrac{1}{2}a{\left( {0.5} \right)^2}$
By performing various arithmetic operations, we get
$\Rightarrow a = 4\,m{s^{ - 2}}$
Using the other equation of motion,
$v = u + at$
Substituting the initial velocity, acceleration and the time taken in it
$\Rightarrow v = 0 + 4 \times 0.5 = 2\,m{s^{ - 1}}$
Hence the final velocity of the body is obtained as $2\,m{s^{ - 1}}$ . For the next $0.5\,m$ , the previous final velocity becomes the initial velocity but the acceleration remains the same.
$s = ut + \dfrac{1}{2}a{t^2}$
$\Rightarrow s = 2t + \dfrac{1}{2}4{t^2}$
By simplifying the above equation, we get
$\Rightarrow 2t + 2{t^2} - 0.5 = 0$
By solving the above equation, we get
$\Rightarrow t = 0.207s$
Hence the time taken to cover the second $0.5\,m$ is $0.21\,s$.
Note: The velocity of the body that slides on the inclined surface mainly depends on the friction of the surface, mass of the body and the slope of the inclination of the surface. If the surface is rough and so the friction is more and hence the velocity of the body reduces. If it has a smooth slippery surface, it has a greater velocity.
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