Motion in a Straight Line- 1 Practice Questions - DPP for NEET

 Page 1


1. ( a) If t
1
 and 2 t
2
 are the time taken by particle to cover first
and second half distance respectively.
/2
36
xx
t==
...(i)
x
1
 = 4.5t
2
 and x
2
 = 7.5 t
2
So, 
1 2 22
4.5 7.5
22
+ =Þ +=
xx
xx tt
2
24
x
t= ...(ii)
Total time t = 
12
2
6 124
x xx
tt + = +=
So,average speed = 4 m/sec.
2. (c)
2
1
2
dv bt
bt dv btdt v K
dt
= Þ = Þ=+
At 
0 10
0, t v v Kv = = Þ=
We get 
2
0
1
2
v btv =+
Again 
2
2
0 02
11
2 23
dx bt
bt v x vtK
dt
= + Þ = ++
At 
2
0,00 t xK = =Þ=
\
3
0
1
6
x bt vt =+
3. (d) 
63
63
dv dv
v dt
dtv
=-Þ=
-
Integrating both sides, 
(6 3)
=
-
òò
dv
dt
v
Þ
1
log (6 3)
3
e
v
tK
-
=+
-
Þ log
e
(6 – 3v) = – 3t + K
2
....(i)
At t = 0, v = 0 Þ log
e
 6 = K
2
Substituting the value of K
2
 in equation (i)
log
e
(6 – 3v) = – 3t + log
e
 6
Þ
3
6 3 63
log3
66
t
e
vv
te
-
-- æö
=-Þ=
ç÷
èø
Þ
33
63 6 3 6(1)
tt
ve ve
--
- = Þ =-
Þ
3
2(1)
t
ve
-
=-
\ v
terminal
 = 2 m/s (when t = ¥)
Acceleration a = 
33
2(1 )6
tt
dvd
ee
dt dt
--
éù
= -=
ëû
Initial acceleration  = 6 m/s
2
.
4. (a) The body starts from rest at x = 0 and then again comes
to rest at x = 1. It means initially acceleration is positive
and then negative.
So we can conclude that a can not remain positive for
all t in the interval 01 t ££ i.e. a must change sign
during the motion.
5. ( b) The area under acceleration time graph gives change
in velocity . As acceleration is zero at the end of 11 sec.
i.e. v
max
= Area of DOAB
1
11 10 55m/s
2
= ´ ´=
11 sec.
t
10 / ms
2
a
A
B
O
6. (d) Let the car accelerate at rate a for time t
1
 then maximum
velocity attained, v = 0 + at
1
 = at
1
Now , the car decelerates at a rate b for time (t – t
1
) and
finally comes to rest. Then,
0 = v – b(t – t
1
)
Þ 0 = at
1
 – bt + bt
1
Þ
1
tt
b
=
a +b
Þvt
ab
=
a+b
7. (c) S
n
 = (2 1)
2
+-
a
un
= (2 1)
2
-
a
n ( 0) = Qu
[ ]
1
2( 1) 1 (2 n 1)
22
+
= +-= +
n
aa
Sn
\
1
21
21
+
-
=
+
n
n
S n
Sn
Page 2


1. ( a) If t
1
 and 2 t
2
 are the time taken by particle to cover first
and second half distance respectively.
/2
36
xx
t==
...(i)
x
1
 = 4.5t
2
 and x
2
 = 7.5 t
2
So, 
1 2 22
4.5 7.5
22
+ =Þ +=
xx
xx tt
2
24
x
t= ...(ii)
Total time t = 
12
2
6 124
x xx
tt + = +=
So,average speed = 4 m/sec.
2. (c)
2
1
2
dv bt
bt dv btdt v K
dt
= Þ = Þ=+
At 
0 10
0, t v v Kv = = Þ=
We get 
2
0
1
2
v btv =+
Again 
2
2
0 02
11
2 23
dx bt
bt v x vtK
dt
= + Þ = ++
At 
2
0,00 t xK = =Þ=
\
3
0
1
6
x bt vt =+
3. (d) 
63
63
dv dv
v dt
dtv
=-Þ=
-
Integrating both sides, 
(6 3)
=
-
òò
dv
dt
v
Þ
1
log (6 3)
3
e
v
tK
-
=+
-
Þ log
e
(6 – 3v) = – 3t + K
2
....(i)
At t = 0, v = 0 Þ log
e
 6 = K
2
Substituting the value of K
2
 in equation (i)
log
e
(6 – 3v) = – 3t + log
e
 6
Þ
3
6 3 63
log3
66
t
e
vv
te
-
-- æö
=-Þ=
ç÷
èø
Þ
33
63 6 3 6(1)
tt
ve ve
--
- = Þ =-
Þ
3
2(1)
t
ve
-
=-
\ v
terminal
 = 2 m/s (when t = ¥)
Acceleration a = 
33
2(1 )6
tt
dvd
ee
dt dt
--
éù
= -=
ëû
Initial acceleration  = 6 m/s
2
.
4. (a) The body starts from rest at x = 0 and then again comes
to rest at x = 1. It means initially acceleration is positive
and then negative.
So we can conclude that a can not remain positive for
all t in the interval 01 t ££ i.e. a must change sign
during the motion.
5. ( b) The area under acceleration time graph gives change
in velocity . As acceleration is zero at the end of 11 sec.
i.e. v
max
= Area of DOAB
1
11 10 55m/s
2
= ´ ´=
11 sec.
t
10 / ms
2
a
A
B
O
6. (d) Let the car accelerate at rate a for time t
1
 then maximum
velocity attained, v = 0 + at
1
 = at
1
Now , the car decelerates at a rate b for time (t – t
1
) and
finally comes to rest. Then,
0 = v – b(t – t
1
)
Þ 0 = at
1
 – bt + bt
1
Þ
1
tt
b
=
a +b
Þvt
ab
=
a+b
7. (c) S
n
 = (2 1)
2
+-
a
un
= (2 1)
2
-
a
n ( 0) = Qu
[ ]
1
2( 1) 1 (2 n 1)
22
+
= +-= +
n
aa
Sn
\
1
21
21
+
-
=
+
n
n
S n
Sn
6
DPP/ P 03
 
. 8 (c) As acc. is constant so from s = ut + 
1
2
 at
2
 we have
x = 
1
2
 at
2
   [u = 0] ....(i)
Now if it travels a distance y in next t sec.
in 2t sec total distance travelled
x + y = 
1
2
 a(2t)
2
....(ii) (t + t = 2t)
Dividing eq
n
. (ii) by eq
n
 (i),  
xy
4
x
+
=  or  y = 3x
. 9 (b)
9
7
) 1 5 2 (
2
g
) 1 4 2 (
2
g
) 5 ( x
) 4 ( x
=
- ´
- ´
=
        th
n
a
[ S u (2n 1)]
2
=+- Q
10 (c) Let body takes T sec to reach maximum height.
Then v = u – gT
v = 0, at highest point.
g
u
T =
 ...................(1]
H
A
C
B v = 0
u
V elocity attained by body in (T – t) sec
v = u – g (T – t)
   = u – gT + gt = u –
gt
g
u
g +
or v = gt ...................(2)
\Distance travelled in last t sec of its ascent
2 2
gt
2
1
gt
2
1
t ) gt ( S = - =
1 1 . (c) v = 
dxd
dt dt
= (2t
2
 + t + 5) =  4t + 1 m/s
and  a = 
dt
d
dt
dv
=
 (4t + 1)   ;  a  =  4  m/s
2
2 1 . (a) 0
dvt
a1
dtT
æö
=-
ç÷
èø
Þ 
vt
2
00
00
tt
dv a 1 dt v a t
T 2T
æö
æö
= - Þ=-
ç÷ ç÷
èø
èø
òò
dx
v
dt
= Q so, 
t
2
0
0
t
dx v dt x a t dt
2T
æö
= Þ=-
ç÷
èø
òòò
Þ 
23
0
tt
xa
2 6T
æö
=-
ç÷
èø
Q a = 0 Þ  t = T
Average velocity  = 
displacement
time
= 
23
0
0
TT
a
2 6T
aT
T3
æö
-
ç÷
èø
=
3 1 . (b) Let u
1
, u
2
, u
3
 and u
4
 be velocities at time t = 0, t
1
,
(t
1
 + t
2
) and (t
1
 + t
2
 + t
3
) respectively and acceleration
is a then
v
1
 = 
23 12
2
,
22
uu uu
v
+ +
= and 
34
3
2
uu
v
+
=
Also u
2
 = u
1
 + at
1
, u
3
 = u
1
 + a(t
1
 + t
2
)
and u
4
 = u
1
 + a(t
1
 + t
2
 + t
3
)
By solving, we get
1 2 12
2 3 23
()
()
-+
=
-+
v v tt
v v tt
4 1 . (c) x
dxd
v
dt dt
==
 (3t
2
 – 6t) = 6t – 6. At t = 1, v
x
 = 0
y
dyd
v
dt dt
==
 (t
2
 – 2t) = 2t – 2. At t = 1, v
y
 = 0
Hence v = 
22
0
xy
vv +=
5 1 . (c) Initial relative velocity = v
1
 – v
2
.
Final relative velocity = 0
From v
2
 = u
2
 – 2as
Þ 0 = (v
1
 – v
2
)
2
 – 2 × a × s
Þ
2
12
()
2
vv
s
a
-
=
If the distance between two cars is 's' then collision will
take place. To avoid collision d > s
\  
2
12
()
2
vv
d
a
-
>
where d = actual initial distance between two cars.
16 (c) If the body starts from rest and moves with constant
acceleration then the ratio of distances in consecutive
equal time interval S
1
 : S
2
 : S
3
 = 1 : 3 : 5
7 1 . (c) Instantaneous velocity  
x
v
t
D
=
D
By using the data from the table
12
0 ( 2) 60
2m/s, 6m/s,
11
vv
---
= = ==
3
166
10m/s
1
v
-
==
So, motion is non-uniform but accelerated.
Page 3


1. ( a) If t
1
 and 2 t
2
 are the time taken by particle to cover first
and second half distance respectively.
/2
36
xx
t==
...(i)
x
1
 = 4.5t
2
 and x
2
 = 7.5 t
2
So, 
1 2 22
4.5 7.5
22
+ =Þ +=
xx
xx tt
2
24
x
t= ...(ii)
Total time t = 
12
2
6 124
x xx
tt + = +=
So,average speed = 4 m/sec.
2. (c)
2
1
2
dv bt
bt dv btdt v K
dt
= Þ = Þ=+
At 
0 10
0, t v v Kv = = Þ=
We get 
2
0
1
2
v btv =+
Again 
2
2
0 02
11
2 23
dx bt
bt v x vtK
dt
= + Þ = ++
At 
2
0,00 t xK = =Þ=
\
3
0
1
6
x bt vt =+
3. (d) 
63
63
dv dv
v dt
dtv
=-Þ=
-
Integrating both sides, 
(6 3)
=
-
òò
dv
dt
v
Þ
1
log (6 3)
3
e
v
tK
-
=+
-
Þ log
e
(6 – 3v) = – 3t + K
2
....(i)
At t = 0, v = 0 Þ log
e
 6 = K
2
Substituting the value of K
2
 in equation (i)
log
e
(6 – 3v) = – 3t + log
e
 6
Þ
3
6 3 63
log3
66
t
e
vv
te
-
-- æö
=-Þ=
ç÷
èø
Þ
33
63 6 3 6(1)
tt
ve ve
--
- = Þ =-
Þ
3
2(1)
t
ve
-
=-
\ v
terminal
 = 2 m/s (when t = ¥)
Acceleration a = 
33
2(1 )6
tt
dvd
ee
dt dt
--
éù
= -=
ëû
Initial acceleration  = 6 m/s
2
.
4. (a) The body starts from rest at x = 0 and then again comes
to rest at x = 1. It means initially acceleration is positive
and then negative.
So we can conclude that a can not remain positive for
all t in the interval 01 t ££ i.e. a must change sign
during the motion.
5. ( b) The area under acceleration time graph gives change
in velocity . As acceleration is zero at the end of 11 sec.
i.e. v
max
= Area of DOAB
1
11 10 55m/s
2
= ´ ´=
11 sec.
t
10 / ms
2
a
A
B
O
6. (d) Let the car accelerate at rate a for time t
1
 then maximum
velocity attained, v = 0 + at
1
 = at
1
Now , the car decelerates at a rate b for time (t – t
1
) and
finally comes to rest. Then,
0 = v – b(t – t
1
)
Þ 0 = at
1
 – bt + bt
1
Þ
1
tt
b
=
a +b
Þvt
ab
=
a+b
7. (c) S
n
 = (2 1)
2
+-
a
un
= (2 1)
2
-
a
n ( 0) = Qu
[ ]
1
2( 1) 1 (2 n 1)
22
+
= +-= +
n
aa
Sn
\
1
21
21
+
-
=
+
n
n
S n
Sn
6
DPP/ P 03
 
. 8 (c) As acc. is constant so from s = ut + 
1
2
 at
2
 we have
x = 
1
2
 at
2
   [u = 0] ....(i)
Now if it travels a distance y in next t sec.
in 2t sec total distance travelled
x + y = 
1
2
 a(2t)
2
....(ii) (t + t = 2t)
Dividing eq
n
. (ii) by eq
n
 (i),  
xy
4
x
+
=  or  y = 3x
. 9 (b)
9
7
) 1 5 2 (
2
g
) 1 4 2 (
2
g
) 5 ( x
) 4 ( x
=
- ´
- ´
=
        th
n
a
[ S u (2n 1)]
2
=+- Q
10 (c) Let body takes T sec to reach maximum height.
Then v = u – gT
v = 0, at highest point.
g
u
T =
 ...................(1]
H
A
C
B v = 0
u
V elocity attained by body in (T – t) sec
v = u – g (T – t)
   = u – gT + gt = u –
gt
g
u
g +
or v = gt ...................(2)
\Distance travelled in last t sec of its ascent
2 2
gt
2
1
gt
2
1
t ) gt ( S = - =
1 1 . (c) v = 
dxd
dt dt
= (2t
2
 + t + 5) =  4t + 1 m/s
and  a = 
dt
d
dt
dv
=
 (4t + 1)   ;  a  =  4  m/s
2
2 1 . (a) 0
dvt
a1
dtT
æö
=-
ç÷
èø
Þ 
vt
2
00
00
tt
dv a 1 dt v a t
T 2T
æö
æö
= - Þ=-
ç÷ ç÷
èø
èø
òò
dx
v
dt
= Q so, 
t
2
0
0
t
dx v dt x a t dt
2T
æö
= Þ=-
ç÷
èø
òòò
Þ 
23
0
tt
xa
2 6T
æö
=-
ç÷
èø
Q a = 0 Þ  t = T
Average velocity  = 
displacement
time
= 
23
0
0
TT
a
2 6T
aT
T3
æö
-
ç÷
èø
=
3 1 . (b) Let u
1
, u
2
, u
3
 and u
4
 be velocities at time t = 0, t
1
,
(t
1
 + t
2
) and (t
1
 + t
2
 + t
3
) respectively and acceleration
is a then
v
1
 = 
23 12
2
,
22
uu uu
v
+ +
= and 
34
3
2
uu
v
+
=
Also u
2
 = u
1
 + at
1
, u
3
 = u
1
 + a(t
1
 + t
2
)
and u
4
 = u
1
 + a(t
1
 + t
2
 + t
3
)
By solving, we get
1 2 12
2 3 23
()
()
-+
=
-+
v v tt
v v tt
4 1 . (c) x
dxd
v
dt dt
==
 (3t
2
 – 6t) = 6t – 6. At t = 1, v
x
 = 0
y
dyd
v
dt dt
==
 (t
2
 – 2t) = 2t – 2. At t = 1, v
y
 = 0
Hence v = 
22
0
xy
vv +=
5 1 . (c) Initial relative velocity = v
1
 – v
2
.
Final relative velocity = 0
From v
2
 = u
2
 – 2as
Þ 0 = (v
1
 – v
2
)
2
 – 2 × a × s
Þ
2
12
()
2
vv
s
a
-
=
If the distance between two cars is 's' then collision will
take place. To avoid collision d > s
\  
2
12
()
2
vv
d
a
-
>
where d = actual initial distance between two cars.
16 (c) If the body starts from rest and moves with constant
acceleration then the ratio of distances in consecutive
equal time interval S
1
 : S
2
 : S
3
 = 1 : 3 : 5
7 1 . (c) Instantaneous velocity  
x
v
t
D
=
D
By using the data from the table
12
0 ( 2) 60
2m/s, 6m/s,
11
vv
---
= = ==
3
166
10m/s
1
v
-
==
So, motion is non-uniform but accelerated.
DPP/ P 03
7
8 1 . (a) 2
1 2
2
v
at vtt
a
= Þ=
9 1 . (b) x = 4(t – 2) + a(t – 2)
2
At t = 0, x = – 8 + 4a = 4a – 8
4 2 ( 2)
dx
v at
dt
= =+-
At t = 0, v = 4 – 4a = 4(1 – a)
But acceleration,  
2
2
2
dx
aa
dt
==
0 2 . (d)
-a b
= +
t t
x ae be
V elocity ()
tt
dxd
v ae be
dt dt
-ab
==+
. ( ) ()
-a b -a b
= -a + b =- a +b
t t tt
ae be a e be
Acceleration = 
().
tt
a e be
-ab
-a -a+bb
22 tt
a e be
-ab
= a +b
Acceleration is positive so velocity goes on increasing
with time.
1 2 . ( d ) (1)    a = 3 sin 4t
Þ
dv
3sin 4t
dt
=
Þ dv 3sin 4t dt c =+
òò
Þ v = 
3
cos4tc
4
-
+
For initial velocity , t = 0
0
3
vC
4
=-+
At particular value of 
3
C
4
= , 
0
v0 =
Therefore, initial velocity may or may not be zero.
(2)  Acceleration = 0
Þ a = 3 sin4t = 0 Þ sin 4t = 0
Þ 4t = np Þ
n
t
4
p
=
where n = 0, 1, 2, .................
Therefore, the acceleration of the particle becomes zero
after each interval of 
4
p
 second.
(3)  As acceleration is sinusoidal function of time, so
particle repeats its path periodically . Thus, the particle
comes to its initial position after sometime (period of
function).
(4)  The particle moves in a straight line path as it
performs S.H.M.
Since (1) & (3) are correct, hence correct answer is (d).
2 2 . (c) For an inertial frame of reference, its acceleration should
be zero. As reference frame attached to the earth i.e. a
rotating or revolving frame is accelerating, therefore, it
will be non-inertial.
Thus (2) & (4) are correct, so correct answer is (c).
3 2 . (c) Average speed
12
12
12
2vv Total distance 2x
xx
Total time v v
vv
= ==
+
+
Average velocity 
Total displacement
Total time
=
Q  It comes back to its initial position
\  Total displacement is zero.
Hence, average velocity is zero.
Sol. For Qs. 4 2 -27. a = sin pt
\  dv 2sin t dt =p
òò
 or 
2
v cos tC =- p+
p
At t = 0, v = 0  
2
C \=
p
 or  
2
v (1 cos t) = -p
p
V elocity is always non-negative, hence particle always moves
along positive x-direction.
\  Distance from time t = 0 to t = t is
t t
2
0
0
2 2 1 22
S (1 cos t) dt t sin t t sin t
æö
= - p = - p = -p
ç÷
èø p ppp
p
ò
Also displacement from time t = 0 to t = 
2
2t2
sint -p
p
p
Distance from time t = t to t = 1s = 
2
p
 meters.
. 4 2 (a) 5 2 . (b) 6 2 . (b)
7 2 . (d) Negative slope of position time graph represents that
the body is moving towards the negative direction and
if the slope of the graph decrease with time then it
represents the decrease in speed i.e. retardation in
motion.
8 2 . (c) As per definition, acceleration is the rate of change of
velocity,
i.e. .
dv
a
dt
=
r
r
If velocity is constant
0,0
dv
a
dt
= \=
r
r
Therefore, if a body has constant velocity it cannot
have non zero acceleration.
. 9 2 (d) The displacement is the shortest distance between
initial and final position. When final position of a body
coincides with its initial position, displacement is zero,
but the distance travelled is not zero.