Question

A body moves with a velocity of $2 m / s$ for $5 s$ , then its velocity uniformly increases to $10 m / s$ for the next $5 s$ . Thereafter, its velocity begins to decrease at a uniform rate until it comes to rest after $10 s$ .

(a) Plot a $v - t$ graph and $d - t$ graph for the motion of the body.

(b) Mark the portions of the graph when the motion is uniform and when it is non-uniform.

(c) From the graph find the total distance moved by the body in

first $2 s$
first $12 s$ and
in the last $10 s$

Accepted Answer

(a)

(b) The motion is uniform from point A to B as the velocity is constant in both the graphs.

(c)

For the calculation of distance for any time lapses, we can simply look the distance-time graph and find the information. We can also use speed-time graph and calculate areas underneath the graph for distances.

From the graph above it can be seen that for the first 2s, speed $= 2 m / s$ .

Therefore, distance covered is the area under the graph, i.e., $(2 m / s) \times (2 s) = 4 m$

In initial $12$ seconds, distance travelled ( $s_{12 s}$ ) is the Area of rectangle $OABM +$ Area of trapezium $BCNM +$ Area of trapezium $NCQP$

$s_{12 s} = (OM \times OA) + \frac{1}{2} (BM + CN) \times MN + \frac{1}{2} (CN + QP) \times NP$

$s_{12 s} = [5 s \times 2 m / s] + \frac{1}{2} (2 m / s + 10 m / s) \times 5 s + \frac{1}{2} (10 m / s + QP) \times 2 s$

Considering motion from $C$ to $Q$ , we can find out the final velocity $v$ at $12 s$ using the formula, $v = u + a t$

From the graph, initial velocity, $u = 10 m / s$ , time taken for the velocity change, $t = 2 s$

We know that the acceleration ( $a$ ) is the slope of the velocity-time graph

$a =$ slope of $CQ$ $=$ slope of $CD = - \frac{10}{10} = - 1 m / s^{2}$ ,

So now, $v = 10 - 1 \times (2) = 8 m / s$

Therefore, distance travelled in $12 s$

$s_{12 s} = [5 s \times 2 m / s] + \frac{1}{2} (2 m / s + 10 m / s) \times 5 s + \frac{1}{2} (10 m / s + 8 m / s) \times 2 s$ $= 58 m$ .

In the last $10$ seconds.

Distance travelled in the last $10 s$ , $s_{10 e n d} =$ Area of $△ NCD$

$s_{10 e n d} = \frac{1}{2} \times (ND) \times (CN)$ $= \frac{1}{2} \times 10 s \times 10 m / s$ $= 50 m$

Sunil Batra Solutions for Chapter: Motion, Exercise 4: HOTS Higher Order Thinking Skill & Brain Twisting

Sunil Batra Physics Solutions for Exercise - Sunil Batra Solutions for Chapter: Motion, Exercise 4: HOTS Higher Order Thinking Skill & Brain Twisting

Questions from Sunil Batra Solutions for Chapter: Motion, Exercise 4: HOTS Higher Order Thinking Skill & Brain Twisting with Hints & Solutions

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