Question

The v-x graph has been provided. Predict the a-x graph.

v - velocity
x - position
a - acceleration

#Revision Q11​

Answer 1

Answer:

We are given with a velocity - displacement graph and we need to predict the acceleration - displacement graph.

So, from the given v-x graph we can see that as x is increasing v is decreasing, therefore we can say that the slope of this v-x graph is Negative. Now, let us assume that when the velocity is zero the particle has travelled a distance of x₀.

Therefore, let's write the equation of graph involved, from the graph we can make out that the graph has an equation of y= mx + c (Where m is the slope of the v-x graph).

Now, let's put the values,

⇒ v = (0-v₀/x₀-0) × x + c

⇒ v = -v₀x/x₀ + c _[1]

Therefore, we got the equation of the graph.

Now, from the relation of acceleration and displacement we know that,

⇒ a = v.dv/dx

Differentiating equation 1 w.r.t dx and substitute in dv/dx,

⇒ a = v × -v₀/x₀

⇒ a = -v.v₀/x₀

Now, substitute the value of equation 1 in the above

⇒ a = v₀/x₀ × -(-v₀/x₀ + c)

⇒ a = v₀/x₀ × (v₀/x₀ - c)

⇒ a = (v₀/x₀)² - v₀.c/x₀

Now, we can see that v₀,c,x₀ are constant and therefore we can say that v₀.c/x₀ = C (Where C is a constant)

⇒ a = (v₀/x₀)² - C

This equation resembles the equation y = mx - c.

Therefore the graph will have a positive slope but the y-intercept will be negative.

So, the a-x graph will be:- (Image is attached)

Answer 2

From given graph:

$v = (slope)x + v_{0}$

Since velocity decreases with increase in displacement, slope will be negative , let it be -k.

$= > v = - kx + v_{0}$

Now, acceleration is :

$a = \frac{dv}{dt}$

$= > a = \frac{dv}{dx} \times \frac{dx}{dt}$

$= > a = v \frac{dv}{dx}$

$= > a = v \times \frac{d( - kx + v_{0})}{dx}$

$= > a = v \times ( - k)$

$= > a = ( - kx + v_{0})\times ( - k)$

$= > a = {k}^{2} x - k v_{0}$

So, negative intercept and positive slope.