The best thinking has been done in solitude. The worst has been done in turmoil

– Thomas Alva Edison

The essential things you need to remember in *one dimensional uniformly accelerated motion* were discussed in the post dated 13^{th} August 2008. A few multiple choice questions on one dimensional motion were discussed in the post dated 21^{st} August 2008. You can access both posts by clicking on the

(1) In the straight line graph AB shown in the adjoining figure time is taken along the x-axis. The quantity taken along the y-axis (which is not shown) in the graph can be

(a) acceleration of a train arriving at a railway station

(b) velocity of a train arriving at a railway station

(c) velocity of a cricket ball thrown towards a batsman

(d) acceleration of a ball thrown vertically downwards

(e) velocity of a ball thrown vertically upwards

When a ball is thrown vertically upwards, its velocity *v *goes on decreasing linearly with time *t *in accordance with the equation,

*v* = *v*_{0} – *gt* where *v*_{0} is the velocity of projection and *g* is the acceleration due to gravity.

At the maximum height the velocity is zero. Thereafter the direction of the velocity is reversed (during the return trip of the ball) and the magnitude of velocity goes on increasing linearly with time in accordance with the equation,

*v* = *v*_{0} + *gt*

Therefore, the graph AB is the velocity-time graph of a body projected vertically upwards. Option (e) is correct.

(2) An iron sphere released from rest at a height *h* falls freely under gravity. It reaches half the height at the instant *t*_{1} and hits the ground at the instant *t*_{2}. if air resistance is negligible, the ratio *t*_{1}/*t*_{2} is

(a) 2^{1/2}^{}

(b) 2^{–1/2}

(c) 2^{–1/4}

(d) 2^{1/4 }

(e) ¼

We have *s *= *v*_{0} *t*+ (½) *at*^{2} where *s* is the displacement in time *t*, *v*_{0} is the initial velocity and *a* is the acceleration.

Since the displacements in times *t*_{1 }and *t*_{2} are *h*/2 and *h* respectively, we have

*h*/2 = 0 + (½) *g t*_{1}^{2} and

*h*= 0 + (½) *g t*_{2}^{2}

Therefore *t*_{1}^{2}/*t*_{2}^{2} = ½ so that *t*_{1}/*t*_{2} = 2^{–1/2}

(3) A rifle can fire bullets with a speed of 1000 ms^{–1}. Assume that air resistance is negligible and the gravitational acceleration is 10 ms^{–2}. In order to hit a target at a horizontal distance of 100 m, the rifle is to be aimed

(a) 5 cm above the target

(b) 10 cm above the target

(c) 10 cm below the target

(d) 5 cm below the target

(e) directly at the target

Since the gravitational force does not affect the horizontal velocity of the bullet, the time *t *taken to cover the horizontal distance of the target is given by

*t =* 100/1000 = 0.1 s

During this time the bullet will fall down by a distance *s* given by

*s* = ½ *gt*^{2} = ½ ×10×(0.1)^{2 }= 0.05 m = 5 cm.^{}

Therefore, in order to hit the target, the rifle is to be aimed at a point 5 cm* above* the target [Option (a)].

(4) When a body is thrown vertically down with a velocity of 4 ms^{–1} from a height *h* it* *reaches the ground with a velocity of 5 ms^{–1}. If the same body is is released from rest from the same height *h*, it* *it will reach the ground with a velocity

(a) 2 ms^{–1}

(b) 2.5 ms^{–1}

(c) 3 ms^{–1}

(d) 3.5 ms^{–1}

(e) 4 ms^{–1}

We can use here the equation of uniformly accelerated motion,

*v*^{2} = *u*^{2} + 2*a*s where *u* is the initial velocity and *v* is the final velocity after a body moving with an acceleration *a *suffers a displacement *s*.

When the body is thrown down with a velocity of 4 ms^{–1} from a height *h* we have

5^{2 }= 4^{2} + 2*gh* …….(i)

where *g* is the acceleration due to gravity.

When the body is released from rest from the same height *h* we have

*v*^{2} = 0 + 2*gh* ……..(ii)

Subtracting Eq.(ii) from Eq.(i) we obtain

5^{2} – *v*^{2} = 4^{2} from which *v* = 3 ms^{–1}.

The following questions [(5) and (6)] are meant for AP Physics C aspirants:

(5) A particle moving along the positive x-direction has its position coordinate *x* at time *t *given by

√*x* = *b *

*t+*

*c*where

*b*and

*c*are constants.

The acceleration and the initial velocity of the particle are respectively

(a) zero, *b *

(b) zero, 2*bc*

(c) *b*^{2}, *bc*

(d) *b*^{2}, 2*bc*

*b*

^{2}, 2

*bc*

Squaring the equation for the position coordinate, we have

*x = b*^{2}*t*^{2} + *c*^{2}* **+ *2*bct*

The velocity *v *of the particle at the instant *t* is given by

*v = dx/dt = *2* b*^{2}*t* + 2*bc*

Or, *v =*2*bc + *2*b*^{2}*t*

When *t = *0, *v = *2*bc*. This is the initial velocity of the particle.

The acceleration *a* of the particle is given by

*a = dv/dt = *2*b*^{2}

[You can obtain the answers from the form of the equation, *x = b*^{2}*t*^{2} + *c*^{2}* + *2*bct* itself. This equation is in the form, *x* = *x*_{0} + *v*_{0}*t + *½ *at*^{2} which gives the displacement (*x* – *x*_{0}) at the instant *t* of a particle moving along a straight line with acceleration *a *and initial velocity *v*_{0}].

(6) A ball projected vertically up from the ground moves past a point P at a height *h *at the end of 2 seconds during the upward trip. After another 2 seconds it moves past the same point P during the downward trip. Neglect air resistance and assume that *g = *10 ms^{–2}. The maximum height reached by the ball is

(a) 50 m

(b) 45 m

(c) 40 m

(d) 33 m

(e) 22 m

The time taken by the ball to reach the maximum height is 2 +1 = 3 s.

If *u* is the velocity of projection and *t* is the time taken to reach the maximum height, we have (from the equation *v = u + at*)

0 = *u *– *gt = u** *– 10×3

This gives *u = *30 ms^{–1}

The maximum height reached is *ut *– ½ *gt*^{2} = 30×3* *– ½ ×10×3^{2} = 45 m.

(7) A car starts from rest and moves along straight road with uniform acceleration *α* for some time. Then it moves with uniform retardation *β* and comes to rest.* *If the total time of travel is *t* what is the maximum velocity attained by the car?

(a) 2α*βt /*(*α+ β *)

(b) (*α+ β *)*t* /α*β*

(c) *αβt /*(*α+ β *)

(d) *αt* */*(*α+ β *)

(e) *α*^{2}*βt /*(*α+ β *)

If *t*_{1}^{ }is the time interval during which the car was *accelerated* and *v* is the *maximum* velocity of the car during the trip we have the following equations considering the accelerated and decelerated parts of the trip respectively:

*v =*0 + α* t*_{1} …………….(i)

0 = *v *–* β*(*t *– *t*_{1}) ………..(ii)

From these equations,* *α* t*_{1} = *β*(*t *– *t*_{1}).

Or, *t*_{1}(*α+ β *) = *βt* from which *t*_{1 }= *βt /*(*α+ β *).

From Eq.(i) the *maximum* velocity *v = *α* t*_{1}.

Substituting for *t*_{1}, the *maximum* velocity *v =* α*βt /*(*α+ β *).

(8) In the above question what is the distance traveled during the *decelerated *part of the trip?

(a) α*β*^{ }*t*^{2}* /*2(*α+ β *)^{2}

(b) α*β*^{ }*t*^{2}* /*2(*α+ β *)

(c) α^{2}*β*^{ }*t*^{2}* /*2(*α+ β *)^{2}

(d) α^{2}*β*^{ }*t /* (*α+ β *)^{2}

(e) α^{2}*β*^{ }*t*^{2}* /* (*α+ β *)^{2}

During the *decelerated *part of the trip the initial velocity is the maximum velocity *v *which is equal to α*βt /*(*α+ β *). The final velocity is zero. Therefore the distance traveled (*s*) during the *decelerated *part of the trip is given by

0 = *v*^{2 }– 2* βs* ^{}

Thus we have *s = v*^{2}* /*2*β* = [α*βt /*(*α+ β *)]^{2 }/2*β* = α^{2}*β*^{ }*t*^{2}* /*2(*α+ β *)^{2}

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