Question

. Figure shows the speed - time graph of a car. Using the graph answer the following questions. i) What is the acceleration of the car in first two hours, in the next two hours and in the last two hours? ii) What is the distance travelled by the car?​

Answer 1

Solution:

Explanation:

→ In a speed time graph, the slope of graph tell us about the acceleration. If the slope is high then the acceleration is positive, if the solve is low then the acceleration is negative and if the slope is parallel to time axis then there is no acceleration taking place!

→ In a speed time graph area under curve tell us about the distance or displacement travelled!

Required solution:

Solution 1st

~ Finding acceleration in first two hours!

Here: The final velocity is 30 km/h, initial velocity is 0 km/h and time taken is 2 hours (∵ 2-0)

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Change \: in \: time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{30-0}{2} \\ \\ :\implies \sf a \: = \dfrac{30}{2} \\ \\ :\implies \sf a \: = 15 \: kmh^{-2} \\ \\ :\implies \sf Acceleration \: = 15 \: kmh^{-2}$

Therefore, acceleration = 15 km/h

~ Now let's find acceleration in next two hours!

Here: the final velocity is 60 km/h, initial velocity is 30 km/h and time taken is 2 hours (∵ 4-2)

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Change \: in \: time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{60-30}{2} \\ \\ :\implies \sf a \: = \dfrac{30}{2} \\ \\ :\implies \sf a \: = 15 \: kmh^{-2} \\ \\ :\implies \sf Acceleration \: = 15 \: kmh^{-2}$

Therefore, acceleration = 15 km/h

~ Now let's find acceleration in last two hours!

Here: Final velocity is 0 km/h, initial velocity is 30 km/h and time is 2 hrs (∵ 8-6)

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Change \: in \: time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{0-30}{2} \\ \\ :\implies \sf a \: = \dfrac{-30}{2} \\ \\ :\implies \sf a \: = -15 \: kmh^{-2} \\ \\ :\implies \sf Acceleration \: = -15 \: kmh^{-2}$

Therefore, acceleration =-15 km/h

Solution 2nd

~ Firstly finding the area of rectangle!

$:\implies \sf Distance \: = Area \: under \: curve \\ \\ :\implies \sf s \: = Area \: under \: curve \\ \\ :\implies \sf s \: = Area \: of \: rectangle \\ \\ :\implies \sf s \: = Length \times Breadth \\ \\ :\implies \sf s \: = (60-0) \times (4-0) \\ \\ :\implies \sf s \: = 60 \times 4 \\ \\ :\implies \sf s \: = 240 \: unit \: sq.$

~ Now finding area of triangle!

$:\implies \sf Distance \: = Area \: under \: curve \\ \\ :\implies \sf s \: = Area \: under \: curve \\ \\ :\implies \sf s \: = Area \: of \: triangle \\ \\ :\implies \sf s \: = \dfrac{1}{2} \: Base \times Height \\ \\ :\implies \sf s \: = \dfrac{1}{2} \times (8-4) \times (60-0) \\ \\ :\implies \sf s \: = \dfrac{1}{2} \times 4 \times 60 \\ \\ :\implies \sf s \: = \dfrac{1}{2} \times 240 \\ \\ :\implies \sf s \: = 120 \: unit \ sq.$

~ Now finding total distance!

$:\implies \sf Distance \: = Area \: of \: rectangle \: + Area \: of \: \triangle \\ \\ :\implies \sf Distance \: = 240 + 120 \\ \\ :\implies \sf Distance \: = 360 \: km$

Therefore, distance = 360 km