Name: Understanding Speed-Time Graph
Uploaded: 2019-07-19
Duration: 2 min 44 s
Description: Any idea, information or concept represented graphically is easier to understand than an equation or other mathematical form. Watch this video to realise this through the speed-time graph.

Question

A body moves with a velocity of $2 {ms}^{- 1}$ for $5 s$ . Then its velocity increases uniformly to $10 {ms}^{- 1}$ in next $5 s$ . Thereafter, its velocity begins to decrease at a uniform rate until it comes to rest after $10 s$ . From the velocity-time graph, find the total distance moved by the body in initial $5 s$ and initial $12 s$ and in the last $10 s$ .

Accepted Answer

The velocity-time graph of the body will be like,

We know that the area of a velocity-time graph gives the distance travelled by the body.

So, in the initial $5$ seconds,

Distance travelled ( $d_{5}$ ) is the area of rectangle $OABM$

$d_{5} = OM \times OA = 5 s \times 2 {ms}^{- 1} = 10 m$

In initial $12$ seconds,

Distance travelled ( $d_{12}$ ) is the Area of rectangle $OABM +$ Area of trapezium $BCNM +$ Area of trapezium $NCQP$

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

$d_{12} = [5 s \times 2 {ms}^{- 1}] + \frac{1}{2} (2 {ms}^{- 1} + 10 {ms}^{- 1}) \times 5 s + \frac{1}{2} (10 {ms}^{- 1} + QP) 2 s$

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, $u = 10 {ms}^{- 1}$

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 $CD = - \frac{10}{10} = - 1 {ms}^{- 2}$ ,

That is, $v = 10 - (1 \times 2) = 8 {ms}^{- 1}$

Therefore, distance travelled in $12 s$

$d_{12} = [5 s \times 2 {ms}^{- 1}] + \frac{1}{2} (2 {ms}^{- 1} + 10 {ms}^{- 1}) \times 5 s + \frac{1}{2} (10 {ms}^{- 1} + 8 {ms}^{- 1}) \times 2 s$ $= 58 m$ .

In the last $10$ seconds.

Distance travelled in the last $10 s$ , $d =$ Area of $△ NCD$

$d = \frac{1}{2} \times (ND) \times (CN)$ $= \frac{1}{2} \times 10 s \times 10 {ms}^{- 1}$ $= 50 m$

Important Questions on Motion

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