Motion in a Straight Line Practice Questions - DPP for JEE

 Page 1


1. (b) x = at
3
 and  y = ßt
3
 and 
2. (d) Let ‘S’ be the distance between two ends and ‘a’ be the constant
acceleration
As we know v
2
 – u
2
 = 2aS
or, aS = 
Let v
c
 be velocity at mid point.
Therefore, 
v
c
 = 
3. (a) Acceleration of the particle a = 2t – 1
The particle retards when acceleration is opposite to velocity.
? a . v < 0 ? (2t – 1) (t
2
 – t) < 0 ? t (2t – 1) (t – 1) < 0
Now t is always positive
?  (2t – 1) (t – 1) < 0
or  2t – 1 < 0  and  t – 1 > 0 ? t <  and t > 1.
This is not possible
or  2t – 1 > 0 and t – 1 < 0 ? 1/2 < t < 1
4. (b) When a ball is dropped on a floor,
Page 2


1. (b) x = at
3
 and  y = ßt
3
 and 
2. (d) Let ‘S’ be the distance between two ends and ‘a’ be the constant
acceleration
As we know v
2
 – u
2
 = 2aS
or, aS = 
Let v
c
 be velocity at mid point.
Therefore, 
v
c
 = 
3. (a) Acceleration of the particle a = 2t – 1
The particle retards when acceleration is opposite to velocity.
? a . v < 0 ? (2t – 1) (t
2
 – t) < 0 ? t (2t – 1) (t – 1) < 0
Now t is always positive
?  (2t – 1) (t – 1) < 0
or  2t – 1 < 0  and  t – 1 > 0 ? t <  and t > 1.
This is not possible
or  2t – 1 > 0 and t – 1 < 0 ? 1/2 < t < 1
4. (b) When a ball is dropped on a floor,
   ..... (1)
So the graph between y and t is a parabola. Here as time increases, y
decreases.
When the ball bounces back, then
    ..... (2)
The graph between y and t will be a parabola. But here as time increases,
y also increases. So (b) represents the graph.
5. (d)
   
velocity ,v
0
 = a
accleration 
At t = 0, and
At , 
At 
It cannot go beyond this, so point  is not reached by the particle.
At t = 0, x = 0, at , therefore the particle does not come
back to its starting point at t = 8.
6. (a) or  
Integrating we get, ...(1)
Page 3


1. (b) x = at
3
 and  y = ßt
3
 and 
2. (d) Let ‘S’ be the distance between two ends and ‘a’ be the constant
acceleration
As we know v
2
 – u
2
 = 2aS
or, aS = 
Let v
c
 be velocity at mid point.
Therefore, 
v
c
 = 
3. (a) Acceleration of the particle a = 2t – 1
The particle retards when acceleration is opposite to velocity.
? a . v < 0 ? (2t – 1) (t
2
 – t) < 0 ? t (2t – 1) (t – 1) < 0
Now t is always positive
?  (2t – 1) (t – 1) < 0
or  2t – 1 < 0  and  t – 1 > 0 ? t <  and t > 1.
This is not possible
or  2t – 1 > 0 and t – 1 < 0 ? 1/2 < t < 1
4. (b) When a ball is dropped on a floor,
   ..... (1)
So the graph between y and t is a parabola. Here as time increases, y
decreases.
When the ball bounces back, then
    ..... (2)
The graph between y and t will be a parabola. But here as time increases,
y also increases. So (b) represents the graph.
5. (d)
   
velocity ,v
0
 = a
accleration 
At t = 0, and
At , 
At 
It cannot go beyond this, so point  is not reached by the particle.
At t = 0, x = 0, at , therefore the particle does not come
back to its starting point at t = 8.
6. (a) or  
Integrating we get, ...(1)
At t = 0, v = v
0
  
Putting in (1)  
or  or 
or 
7. (d) Distance from A to B = S = 
Distance from B to C = 
Distance from C to D =  
  
  ............. (i)
............ (ii)
Dividing (i) by (ii), we get  = 
 
8. (a) Q h =  gt
2
Page 4


1. (b) x = at
3
 and  y = ßt
3
 and 
2. (d) Let ‘S’ be the distance between two ends and ‘a’ be the constant
acceleration
As we know v
2
 – u
2
 = 2aS
or, aS = 
Let v
c
 be velocity at mid point.
Therefore, 
v
c
 = 
3. (a) Acceleration of the particle a = 2t – 1
The particle retards when acceleration is opposite to velocity.
? a . v < 0 ? (2t – 1) (t
2
 – t) < 0 ? t (2t – 1) (t – 1) < 0
Now t is always positive
?  (2t – 1) (t – 1) < 0
or  2t – 1 < 0  and  t – 1 > 0 ? t <  and t > 1.
This is not possible
or  2t – 1 > 0 and t – 1 < 0 ? 1/2 < t < 1
4. (b) When a ball is dropped on a floor,
   ..... (1)
So the graph between y and t is a parabola. Here as time increases, y
decreases.
When the ball bounces back, then
    ..... (2)
The graph between y and t will be a parabola. But here as time increases,
y also increases. So (b) represents the graph.
5. (d)
   
velocity ,v
0
 = a
accleration 
At t = 0, and
At , 
At 
It cannot go beyond this, so point  is not reached by the particle.
At t = 0, x = 0, at , therefore the particle does not come
back to its starting point at t = 8.
6. (a) or  
Integrating we get, ...(1)
At t = 0, v = v
0
  
Putting in (1)  
or  or 
or 
7. (d) Distance from A to B = S = 
Distance from B to C = 
Distance from C to D =  
  
  ............. (i)
............ (ii)
Dividing (i) by (ii), we get  = 
 
8. (a) Q h =  gt
2
? h
1
 = g(5)
2
 = 125
h
1
 + h
2
 =  g(10)
2
 = 500
? h
2
 = 375
h
1
  + h
2
 + h
3
 = g(15)
2
 = 1125
? h
3
 = 625
h
2
 = 3h
1
 , h
3
 = 5h
1
or h
1
 =  = 
9. (a) x
A
 = x
B
10.5 + 10t =  ? a = tan 45° = 1 m/s
2
t
2
 – 20t – 21 = 0  ? 
10. (b)
, , 
11. (b) u = 0, t
1
=10s, t
2
 = 20s
Using the relation, S  = ut + at
2
Acceleration being the same in two cases,
Page 5


1. (b) x = at
3
 and  y = ßt
3
 and 
2. (d) Let ‘S’ be the distance between two ends and ‘a’ be the constant
acceleration
As we know v
2
 – u
2
 = 2aS
or, aS = 
Let v
c
 be velocity at mid point.
Therefore, 
v
c
 = 
3. (a) Acceleration of the particle a = 2t – 1
The particle retards when acceleration is opposite to velocity.
? a . v < 0 ? (2t – 1) (t
2
 – t) < 0 ? t (2t – 1) (t – 1) < 0
Now t is always positive
?  (2t – 1) (t – 1) < 0
or  2t – 1 < 0  and  t – 1 > 0 ? t <  and t > 1.
This is not possible
or  2t – 1 > 0 and t – 1 < 0 ? 1/2 < t < 1
4. (b) When a ball is dropped on a floor,
   ..... (1)
So the graph between y and t is a parabola. Here as time increases, y
decreases.
When the ball bounces back, then
    ..... (2)
The graph between y and t will be a parabola. But here as time increases,
y also increases. So (b) represents the graph.
5. (d)
   
velocity ,v
0
 = a
accleration 
At t = 0, and
At , 
At 
It cannot go beyond this, so point  is not reached by the particle.
At t = 0, x = 0, at , therefore the particle does not come
back to its starting point at t = 8.
6. (a) or  
Integrating we get, ...(1)
At t = 0, v = v
0
  
Putting in (1)  
or  or 
or 
7. (d) Distance from A to B = S = 
Distance from B to C = 
Distance from C to D =  
  
  ............. (i)
............ (ii)
Dividing (i) by (ii), we get  = 
 
8. (a) Q h =  gt
2
? h
1
 = g(5)
2
 = 125
h
1
 + h
2
 =  g(10)
2
 = 500
? h
2
 = 375
h
1
  + h
2
 + h
3
 = g(15)
2
 = 1125
? h
3
 = 625
h
2
 = 3h
1
 , h
3
 = 5h
1
or h
1
 =  = 
9. (a) x
A
 = x
B
10.5 + 10t =  ? a = tan 45° = 1 m/s
2
t
2
 – 20t – 21 = 0  ? 
10. (b)
, , 
11. (b) u = 0, t
1
=10s, t
2
 = 20s
Using the relation, S  = ut + at
2
Acceleration being the same in two cases,
?
  S
2
 = 4S
1
12. (a) Velocity of boat 
Velocity of water 
 minutes
13. (c) Here,  or, 
or, 
or, 
where C is the constant of integration.
At t = 0, v = 0.
If f = 0, then
Hence, particle’s velocity in the time interval t = 0 and t = T is given by