
A particle moving in one dimension with a constant acceleration of \[a = 2m/{s^2}\] is observed to cover a distance of \[5\]m during a particular interval of\[ 1\]s. The distance covered by the particle in the next \[1\]s interval is in metre:
A)\[5\]
B)\[6\]
C)\[7\]
D)\[10\]
Answer
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Hint:Acceleration is a vector quantity which tells about magnitude and direction. It is the rate of change of velocity with respect to time. Constant acceleration means that the moving object or particle is changing its velocity by a constant interval of time. The objects falling freely under the effect of gravity often have a constant acceleration. Acceleration is the result of unbalanced force and the net force gives the direction of acceleration.
Step-by-Step Explanation:
Step I: Given acceleration, \[a = 2m/{s^2}\]
In 1 sec, the distance covered is, \[S = 5m\]
Let ‘u’ be the initial velocity.
Step II: Using distance formula,\[S = ut + \dfrac{1}{2}a{t^2}\]
Put values of S, a and t, and find the value of u.
\[5 = (u \times 1) + \dfrac{1}{2} \times 2 \times {(1)^2}\]
\[5 = u + 1\]
\[u = 5 - 1\]
\[u = 4m/s\]
Initial Velocity of the object is, u=4m/s.
Step III: If Distance covered in next one second is to be calculated, then value of \[t = 1 + 1\]
Value of \[t = 2\sec \]
Again using distance formula,\[S = ut + \dfrac{1}{2}a{t^2}\]
Substituting values of u, a and t; find value of S.
\[S = (4 \times 2) + \dfrac{1}{2} \times 2 \times {(2)^2}\]
\[S = 8 + 4\]
\[S = 12m\]
Step IV: Distance covered in \[1st\]second, \[{S_1} = 5m\]
Distance covered in two seconds, \[{S_2} = 12m\]
Distance covered in next one second, \[S = {S_2} - {S_1}\]
\[S = 12 - 5\]
\[S = 7m\]
Distance covered in the interval of the next one second is =7m.
Hence, option C is the right answer.
Note:One dimensional motion of an object means that the object is moving only in one direction. In one dimension the object moves in a straight line. The motion of an object is generally represented in a direction of positive x-axis. Some daily life examples of one dimensional motion includes; an athlete running on a straight racing track, or a person walking straight on zebra crossing etc.
Step-by-Step Explanation:
Step I: Given acceleration, \[a = 2m/{s^2}\]
In 1 sec, the distance covered is, \[S = 5m\]
Let ‘u’ be the initial velocity.
Step II: Using distance formula,\[S = ut + \dfrac{1}{2}a{t^2}\]
Put values of S, a and t, and find the value of u.
\[5 = (u \times 1) + \dfrac{1}{2} \times 2 \times {(1)^2}\]
\[5 = u + 1\]
\[u = 5 - 1\]
\[u = 4m/s\]
Initial Velocity of the object is, u=4m/s.
Step III: If Distance covered in next one second is to be calculated, then value of \[t = 1 + 1\]
Value of \[t = 2\sec \]
Again using distance formula,\[S = ut + \dfrac{1}{2}a{t^2}\]
Substituting values of u, a and t; find value of S.
\[S = (4 \times 2) + \dfrac{1}{2} \times 2 \times {(2)^2}\]
\[S = 8 + 4\]
\[S = 12m\]
Step IV: Distance covered in \[1st\]second, \[{S_1} = 5m\]
Distance covered in two seconds, \[{S_2} = 12m\]
Distance covered in next one second, \[S = {S_2} - {S_1}\]
\[S = 12 - 5\]
\[S = 7m\]
Distance covered in the interval of the next one second is =7m.
Hence, option C is the right answer.
Note:One dimensional motion of an object means that the object is moving only in one direction. In one dimension the object moves in a straight line. The motion of an object is generally represented in a direction of positive x-axis. Some daily life examples of one dimensional motion includes; an athlete running on a straight racing track, or a person walking straight on zebra crossing etc.
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