The distance travelled by a particle starting from rest is plotted against the square of the time elapsed from the commencement of motion. The resulting graph is linear, the slope of this graph is a measure of
initial displacement
initial velocity
acceleration
half the acceleration
half the initial velocity
Explanation
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IT IS C ACCELERATION BECAUSE THE UNIT OF ACCELERATION IS USUALLY KNOWN BY m/s square and the question stated that the graph was been plotted @ the square of time
The answer is D half the acceleration
S = ut + 1/2at^2
Plotting a against t^2 we give is a parabola in which the slope is 1/2a from
Y = MX +C
We are given that the distance traveled by a particle from rest is plotted against the square of the time elapsed, and the resulting graph is linear.
Step 1: Consider the equation of motion
For a particle starting from rest () and moving with constant acceleration (), the distance traveled in time is given by the kinematic equation:
s = ut + \frac{1}{2} a t^2
Since the initial velocity , this simplifies to:
s = \frac{1}{2} a t^2
Step 2: Identifying the slope
If we plot (distance) on the y-axis and (square of time) on the x-axis, the equation takes the form:
s = \left( \frac{1}{2} a \right) t^2
This equation is of the form , where is the slope of the straight-line graph.
Comparing, we see that the slope of the graph is:
\frac{1}{2} a
Step 3: Selecting the correct option
The slope of the graph represents half the acceleration.
Thus, the correct answer is:
D. half the acceleration.
Don't forget that distance(s) is not just equal to velocity multiply by time in this contest because there is not only one velocity, there is more than one velocities(variable velocities).
(Assuming there is no acceleration, velocity is constant and have one value, then we can say distance (s)= velocity × time l. But since there is acceleration, velocity is variable or have different values then distance(s) = average velocity × time = (final velocity(v) - initial velocity(u)) ÷ 2 × time(t).
From s = (1/2)×(v+u)×t. in the question u=0. Therefore s = (1/2)×(v+0)×t which can also be a = (1/2)×(v-0)×t mathematically, since u=0, 0=u and we can have s = (1/2)×(v-u)×t = (1/2)×(v-u)/t×t^2. But acceleration(a) = (v-u)/t. Therefore s = (1/2)×a×t^2. From here it can be deduce that (1/2)×a = s/(t^2). We all know that slope of a graph of distance against time square = s/(t^2) = (1/2)×a.
So, the final answer to the question is the slope of distance against time square graph is half acceleration and not one acceleration. (D)
the answer is C, because from one of the formulas of linear motion s=ut +1/2at² .... make the a which is the acceleration the subject of the formula you will get 2s/t²=a which rhymes with the outcome of the graph being linear and also rhymes with the time which is said to be square of the time ... thank you!!
I think the question is asking for what the slope is a measure of, not what it actually represents. That's probably the only reason the answer is C
