Question

The given figure is an $x - t$ graph of the motion of a particle.

(i) At which of the points $P, Q, R$ and $S$ , the velocity is $v_{x}$ positive?

(ii) At which points the velocity is negative?
(iii) At which points the velocity is zero?

(iv) At which of the points $P, Q, R$ and $S$ is the acceleration $a_{x}$ positive?

(v) At which point is the $a_{x}$ negative?
(vi) At which points does the acceleration appear to be zero?
(vii) At each point, state whether the speed is increasing, decreasing, or not changing.

Accepted Answer

The slope of the tangent at a point in $x - t$ curve is equal to the velocity of the particle at that point, $v_{x} = \frac{d x}{d t}$ .

If the slope is positive, the $v_{x}$ will be positive. If slope is negative then $v_{x}$ is negative and if the slope is zero, then velocity is also zero. The slope of the curve is as shown in diagram below.

So, from the above diagram,

(i) At point $P$ , velocity is positive.

(ii) At point $R$ , velocity is negative.

(iii) At point $Q$ and $S$ the velocity is zero.

The acceleration is actually, rate of change of velocity, so in $x - t$ curve, when it is concave up, so acceleration is positive, for concave down acceleration is negative and for constant slope, the acceleration will be zero. So,

(iv) At point $S$ , the graph is concave up, so acceleration is positive at $S$ .

(v) At point $R$ , the slope is not changing, so the acceleration will be zero at that point.

(vi) From point $P$ to $Q$ , the magnitude of slope is decreasing, so the speed will be decreasing, at point $R$ the speed is constant and at $S$ the slope is zero, so speed is also zero.

B M Sharma Solutions for Chapter: Kinematics I, Exercise 56: ILLUSTRATION 4.52

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Kinematics I, Exercise 56: ILLUSTRATION 4.52

Questions from B M Sharma Solutions for Chapter: Kinematics I, Exercise 56: ILLUSTRATION 4.52 with Hints & Solutions

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