Answer
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Hint: Divide the above journey of the car into 3 parts. The first part from 0 – 20 seconds, the second part from 20 – 90 seconds, and the third part from 90 – 120 seconds. For part (a), observe the graph from 0 – 20 seconds and apply the formula: -
\[a=\dfrac{v-u}{t}\], where ‘a’ is acceleration, ‘v’ is velocity at t = 20s and ‘u’ is velocity at t = 0s. ‘t’ is the time interval of 20 seconds. For part (b), calculate the area under the speed-time graph to find the total distance. For part (c), divide the distance obtained in part (b) with t = 120s.
Complete step-by-step solution
(a) We have to calculate the deceleration of the car during the first 20 seconds.
The term ‘deceleration’ means negative acceleration.
Applying the formula: - \[a=\dfrac{v-u}{t}\], where a = acceleration, v = velocity at t = 20s, u = velocity at t = 0s and t = time interval of 20s, we get,
\[\begin{align}
& \Rightarrow a=\dfrac{10-18}{20}=\dfrac{-8}{20} \\
& \Rightarrow a=-0.4m{{s}^{-2}} \\
\end{align}\]
Since, ‘a’ is negative, therefore deceleration = \[-0.4m{{s}^{-2}}\].
(b) Here, we have to find the total distance traveled by car during the 120 seconds. Therefore, Total distance traveled = Area under the speed-time graph.
\[\Rightarrow \] Total distance travelled
= Area of trapezium ABEDC + Area of rectangle CDGF +Area of trapezium FGIH
Applying the formula of area of trapezium = \[\dfrac{1}{2}\times \] [Sum of parallel sides] \[\times \] height
We have, total distance travelled
\[\begin{align}
& =\dfrac{1}{2}\times \left[ 18+10 \right]\times 20+70\times 10+\dfrac{1}{2}\times \left[ 20+10 \right]\times 30 \\
& =280+700+450 \\
\end{align}\]
= 1430m
Therefore, total distance travelled = 1430m.
(c) Here, we have to find the average speed during the 120 seconds journey.
Applying the formula for average speed = $\dfrac{\text{Total distance}}{\text{Total time}}$
We get, Average speed = \[\dfrac{1430}{120}\].
\[\Rightarrow \] Average speed \[\approx 11.92m{{s}^{-1}}\].
Hence, average speed during the 120s journey is nearly \[11.92m{{s}^{-1}}\].
Note: One may note that we have divided the journey into 3 parts because then it will be easy for us to calculate the area of the speed-time graph which is the total distance traveled. We can also calculate the area in different ways like by forming two triangles and a rectangle. The other method can include the method of integration which is used for complex curves, so we do not need such an integration method here.
\[a=\dfrac{v-u}{t}\], where ‘a’ is acceleration, ‘v’ is velocity at t = 20s and ‘u’ is velocity at t = 0s. ‘t’ is the time interval of 20 seconds. For part (b), calculate the area under the speed-time graph to find the total distance. For part (c), divide the distance obtained in part (b) with t = 120s.
Complete step-by-step solution
(a) We have to calculate the deceleration of the car during the first 20 seconds.
The term ‘deceleration’ means negative acceleration.
Applying the formula: - \[a=\dfrac{v-u}{t}\], where a = acceleration, v = velocity at t = 20s, u = velocity at t = 0s and t = time interval of 20s, we get,
\[\begin{align}
& \Rightarrow a=\dfrac{10-18}{20}=\dfrac{-8}{20} \\
& \Rightarrow a=-0.4m{{s}^{-2}} \\
\end{align}\]
Since, ‘a’ is negative, therefore deceleration = \[-0.4m{{s}^{-2}}\].
(b) Here, we have to find the total distance traveled by car during the 120 seconds. Therefore, Total distance traveled = Area under the speed-time graph.
\[\Rightarrow \] Total distance travelled
= Area of trapezium ABEDC + Area of rectangle CDGF +Area of trapezium FGIH
Applying the formula of area of trapezium = \[\dfrac{1}{2}\times \] [Sum of parallel sides] \[\times \] height
We have, total distance travelled
\[\begin{align}
& =\dfrac{1}{2}\times \left[ 18+10 \right]\times 20+70\times 10+\dfrac{1}{2}\times \left[ 20+10 \right]\times 30 \\
& =280+700+450 \\
\end{align}\]
= 1430m
Therefore, total distance travelled = 1430m.
(c) Here, we have to find the average speed during the 120 seconds journey.
Applying the formula for average speed = $\dfrac{\text{Total distance}}{\text{Total time}}$
We get, Average speed = \[\dfrac{1430}{120}\].
\[\Rightarrow \] Average speed \[\approx 11.92m{{s}^{-1}}\].
Hence, average speed during the 120s journey is nearly \[11.92m{{s}^{-1}}\].
Note: One may note that we have divided the journey into 3 parts because then it will be easy for us to calculate the area of the speed-time graph which is the total distance traveled. We can also calculate the area in different ways like by forming two triangles and a rectangle. The other method can include the method of integration which is used for complex curves, so we do not need such an integration method here.
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