If the displacement of a car is proportional to the square of time, then the car is moving with
Uniform acceleration
Uniform velocity
Increasing acceleration
Decreasing acceleration
Explanation
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Let's break it down:
Since the displacement (s) is proportional to the square of time (t²), we can write the equation:
s = kt²
where k is a constant.
To find the velocity (v) and acceleration (a), we can take the first and second derivatives of the displacement equation with respect to time:
v = ds/dt = 2kt
a = dv/dt = 2k
Notice that the acceleration (a) is constant, since it doesn't depend on time.
This means that the car is moving with:
1. *Uniformly accelerated motion*: The acceleration is constant, but the velocity is changing.
2. *Non-uniform motion*: The velocity is changing, so the motion is not uniform.
In summary, the car's displacement being proportional to the square of time indicates that it's moving with uniformly accelerated motion, where the acceleration is constant, but the velocity is changing.
Let me explain:
the question says displacement /time² which is d/t²
acceleration = velocity /time
velocity = displacement/time
which means displacement/time²
If displacement (s) is proportional to the square of time (t²), it means:
s \propto t^2
Or mathematically,
s = k t^2,k is some constant.Now,
s = ut + \frac{1}{2} a t^2
If the displacement is purely proportional to t², then the initial velocity u must be zero, meaning the car started from rest. The equation then simplifies to
s = \frac{1}{2} a t^2
Comparing this with s = k t², we see that k = ½ a, meaning acceleration a is a constant.
The car is moving with uniform acceleration (constant acceleration). It ain't slowing down or speeding up at an increasing rate—it’s just getting faster at a steady rate every second.
