Motion in a Straight Line

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When the ball dropped from the building u₁=0, u₂=40m/s

Velocity of the dropped ball after time t

V₁=u₁+gt

V₁ = gt

For ball thrown up u₂=40m/s

Velocity of the ball after time t

V₂=u₂-gt

=40-gt

Relative velocity =v₁-v₂

=gt- [-40-gt]=40m/s

This is a short answer type question as classified in NCERT Exemplar

speed of first car = 18km/h

Speed of second car = 27km/hr relative with respect to each other is 18+27=45km/h

Distance between cars = 36km

Time = $\frac{distancebetweencars}{relativespeed}=$ 36/45=0.8h

Distance covered by bird = 36 (0.8)=28.8km

This is a short answer type question as classified in NCERT Exemplar

(a) x (t)=x₀ (1- $e^{-\gamma t}$ )

V (t)=dx/dt= x (t)=x_{0 $\gamma$} ( $e^{-\gamma t}$ )

A (t)= x ( $\infty ,$ t)=x₀ ( ${\gamma }^2{e}^{-\gamma t}$ )

(b) when t=o x (t)= x (t)=x₀ (1- $e^{-0}$ )= x (t)=x₀ (1-1)=0

x (t) is maximum when t= $\infty$

x (t) is maximum when t= $0$

v (t) is maximum when t= $0$ , v (0)=x_{0 $\gamma$}

v (t) is maximum when t= $\infty ,$  v ( $\infty$ )=0

a (t) is maximum when t= $\infty$ , a ( $\infty$ )=0

a (t) is maximum when t= $0$ , a ( $0$ )=-x_{0 $\gamma$} ²

This is a short answer type question as classified in NCERT Exemplar

It is clear from the graph that displacement x is positive throughout . ball is dropped from height and its velocity increases in downward direction due to gravity pull. In this condition v isnegative but acceleration of the ball equal to acceleration due to gravity.

velocity time graph

Acceleration time graph

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When speed becomes constant then acceleration will be zero

a=dv/dt=0

But here a=g-bv

Clearly from above equation as speed increases acceleration will decrease . at a certain speed say v_o, acceleration will be zero and speed will remain constant.

Hence So g=bv

so v= g/b

This is a short answer type question as classified in NCERT Exemplar

x (t)=A+B

Let A>B and

Now velocity is equal to x (t)=dx/dt=-B

So a (t)= dv/dt= B

Above condition are satisfied by the equation.

This is a short answer type question as classified in NCERT Exemplar

(a) The equation we use here is x= 1-sint

So velocity = dx/dt=1-cost

Acceleration =sint

When t=o x=o

When t=, x=

When t=0.x= 2

(b) x=sint so velocity becomes v=cost as displacement and velocity contain sin and cos so equation is periodic.

This is a short answer type question as classified in NCERT Exemplar

If effect of gravity is neglected then ball moving uniformly turned back with same speed when a ball hit it. Acceleration of the ball is zero just before it hits the bat and due  to the repulsive force it gets accelerated.

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We have to analyse slope of each curve i.e= dx/dt . for peak values dx/dt will be zero as x is maximum at peak points.

For graph (a) there is appoint for which displacement is zero so a matches with (iii)

In graph b, x is positive throughout and at point B, V=dx/dt=0

Since at point of curvature changes a=0, so b matches with (ii)

displacement is zero in only first graph so it matches with the (iii) point.

And slope of d graph v=dx/dt is positive so v>0 so acceleration will be negative so matches with I but in graph c it matches with iv as its slope is negative.