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This is one of our many quizzes on Velocity-time graphs. Click on this to access more such competitive quizzes on Mechanics.

Velocity-time graphs Quiz 2

physics

Find the total distance travelled by a cyclist in a race, the variation of speed during his journey is shown in the graph below, $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/0f520175-9087-4681-9b8d-f302fc12dcb33133170592260037126.png"/> 

Velocity is given by dividing distance travelled by time and in a velocity-time graph the x-axis gives the time and y-axis gives the velocity. The area under the graph of a $V$ vs $T$ graph gives the total distance travelled. 

 Velocity is given by dividing distance travelled by time and in a velocity-time graph the x-axis gives the time and y-axis gives the velocity. The area under the graph of a V vs T graph gives the total distance travelled. So, to calculate the area, we divide the figure into parts as shown in the figure below. The area under line AB is a triangle, under line BC is a rectangle, under line CD is the sum of the area of a right-angled triangle and a rectangle and under line DE is a right-angled triangle. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/be8deff1-f9c5-4dd4-aa28-b10b76b0f1b52009058350610975595.png"/> $ \\ $ Therefore, total distance travelled is given by, $ \\ S=A{{r}_{AB}}+A{{r}_{BC}}+A{{r}_{CD}}+A{{r}_{DE}} \\ \Rightarrow S=\left(\dfrac{1}{2}\times 15\times 15\right)+\left(10\times 15\right)+\left[\left(\dfrac{1}{2}\times 10\times 10\right)+\left(10\times 15\right)\right]+\left(\dfrac{1}{2}\times 25\times 10\right) \\ \therefore S=437.5m$ 

Which of the following $V$ vs. $T$ graphs do not show acceleration?

 The slope of the velocity-time graph gives the acceleration of the object in motion. A straight line slope means that the acceleration is uniform and a curved slope shows non-uniform acceleration and when the graph is parallel to the x-axis, meaning there is no change in velocity with time then that body doesn’t have any acceleration. 

 The slope of the velocity-time graph gives the acceleration of the object in motion. A straight line slope means that the acceleration is uniform and a curved slope shows variation in acceleration during motion i.e. non-uniform acceleration and for a straight line slope, if the graph has a positive slope then it depicts acceleration and a negative slope then it depicts negative acceleration i.e. retardation. When the graph is parallel to the x-axis, meaning there is no change in velocity with time then that body doesn’t have any acceleration.

The graph below shows the motion of a bee, what is the average acceleration of the bee during $40$ seconds?  $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/61d2fabf-d539-4951-a0a0-32a37ae917073748932138684488207.png"/>

To find the average acceleration of the bee during $40$ seconds, we draw a line from where the graph is at time $0\,s$ to where the graph is at time $40\,s$ and the gradient of this line is the average acceleration. 

The slope of the velocity-time graph gives the acceleration of the object in motion. A curved slope shows variation in acceleration during motion i.e. non-uniform acceleration. To find the average acceleration of the bee during $40$ seconds, we draw a line from where the graph is at time $0\,s$ to where the graph is at time $40\,s$, as shown in the figure below, $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/e41bd9b6-a529-4103-afca-995f4720c5618258966021663123372.png"/> $ \\ $ So, the average acceleration is given by the gradient of the line joining the origin and point A, $ \\ {{a}_{avg}}=\dfrac{30-0}{40-0} \\ \therefore{{a}_{avg}}=0.75\,m/{{s}^{2}}$

Which course of the journey shown below has maximum acceleration? $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/b8b13625-5466-4f8e-ade7-ae95590341683806729825247073153.png"/> 

For a straight line slope, if the graph has a positive slope then it depicts acceleration and a negative slope then it depicts negative acceleration i.e. retardation and the steeper is the slope of the line, the greater is the acceleration and vice-versa. 

The slope of the velocity-time graph gives the acceleration of the object in motion. A straight line slope means that the acceleration is uniform and for a straight line slope, if the graph has a positive slope then it depicts acceleration and a negative slope then it depicts negative acceleration i.e. retardation. The steeper the slope of the line, the greater is the acceleration and vice-versa. Since, only lines AB and CD show acceleration and AB has a steeper slope than CD, therefore the course of the journey which shows the maximum acceleration is AB. 

 The velocity-time graph of a car is shown below. Find the instantaneous acceleration of the car after it has travelled for $15$ seconds? $ \\ $  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/0a99f4f5-13d3-4744-9937-561d384ff5cd1258635502785051295.png"/>

A curved slope shows variation in acceleration during motion i.e. non-uniform acceleration and to find instantaneous acceleration, we draw a tangent to the curve after $15$ seconds and find the slope of this tangent. 

The slope of the velocity-time graph gives the acceleration of the object in motion. A curved slope shows variation in acceleration during motion i.e. non-uniform acceleration.For instantaneous acceleration, we draw a tangent to the curve after $15$ seconds and find the slope of this tangent, as shown in the figure below, $ \\ $  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/4a5edf17-f64a-4ae5-870a-1c56a99bab3e6879284100830782715.png"/> $ \\ $ Then, instantaneous acceleration is given by, $ \\ a=\dfrac{{{v}_{2}}-{{v}_{1}}}{{{t}_{2}}-{{t}_{1}}} \\ \Rightarrow a =\dfrac{80-40}{25-5} \\ \therefore a =2\,m/{{s}^{2}}$