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If the signs of the velocity and acceleration of a particle are the same, the speed of the particle ______________.- a)Remains same
- b)Increases
- c)Decreases
- d)Cannot conclude anything
Correct answer is option 'B'. Can you explain this answer?
If the signs of the velocity and acceleration of a particle are the same, the speed of the particle ______________.
a)
Remains same
b)
Increases
c)
Decreases
d)
Cannot conclude anything
|
Pooja Shah answered |
In this question, it has been asked about the signs of velocity and acceleration not about the values of velocity and acceleration.
So, if both the entities have same the signs, then the speed of the object will increase.
If it is accelerating in the positive direction, the negative velocity is decreasing, and the object is slowing down.
On acceleration-time graph, a horizontal line indicates:- a)zero displacement
- b)zero acceleration
- c)constant velocity
- d)constant acceleration
Correct answer is option 'D'. Can you explain this answer?
On acceleration-time graph, a horizontal line indicates:
a)
zero displacement
b)
zero acceleration
c)
constant velocity
d)
constant acceleration
|
Anjana Sharma answered |
A horizontal line on a acceleration- time graph indicates that object has constant acceleration.
The dimensions of (change in velocity / time) is- a)[LT-2]
- b)[LT-1]
- c)[L/T-1]
- d)[L/T-2]
Correct answer is option 'A'. Can you explain this answer?
The dimensions of (change in velocity / time) is
a)
[LT-2]
b)
[LT-1]
c)
[L/T-1]
d)
[L/T-2]
|
Naina Sharma answered |
We know that change in velocity / time is acceleration.
And dimension of acceleration is LT-2
A boy throws up a ball in a stationary lift and the ball returns to his hands in 10 s. Now if the lift starts moving up at a speed of 5 m/s. The time taken for a ball thrown straight up to return to his hands is:- a)more than 10 s
- b)Less than 10 s
- c)insufficient information given
- d)equal to 10 s
Correct answer is option 'D'. Can you explain this answer?
A boy throws up a ball in a stationary lift and the ball returns to his hands in 10 s. Now if the lift starts moving up at a speed of 5 m/s. The time taken for a ball thrown straight up to return to his hands is:
a)
more than 10 s
b)
Less than 10 s
c)
insufficient information given
d)
equal to 10 s
|
Suresh Iyer answered |
Lift moving with Uniform speed = 5m / s
Ball is thrown inside the lift
The uniform velocity of the lift does not affect the relative velocity of the ball with respect to the boy.
Relative velocity of ball = 49m/s
⇒ Relative acceleration = 9.8
Thus, u(relative) = velocity of bell
t = 2ur / gr
= (2 × 49) / 9.8
⇒ t = 10s
so, the ball will still return in 10 secs
The dimensions of instantaneous acceleration is:- a)[ LT-2]
- b)[L/T2]
- c)[L/T-1]
- d)[LT-1]
Correct answer is option 'A'. Can you explain this answer?
The dimensions of instantaneous acceleration is:
a)
[ LT-2]
b)
[L/T2]
c)
[L/T-1]
d)
[LT-1]
|
|
Neha Joshi answered |
Instantaneous acceleration is rate of change of instantaneous velocity with time or velocity upon total time, hence we get its dimension is same as of acceleration i.e. LT-2
What does this graph indicate about the motion of an object?
- a)Object is moving in positive direction with positive acceleration
- b)Object is moving in negative direction with negative acceleration
- c)An object is moving in positive direction till time t1, and then turns back with the same Velocity
- d)Object is moving in positive direction with negative acceleration
Correct answer is option 'D'. Can you explain this answer?
What does this graph indicate about the motion of an object?
a)
Object is moving in positive direction with positive acceleration
b)
Object is moving in negative direction with negative acceleration
c)
An object is moving in positive direction till time t1, and then turns back with the same Velocity
d)
Object is moving in positive direction with negative acceleration
|
Top Rankers answered |
Through the given graph we get that:
Velocity is decreasing from a positive constant to some negative number with zero at t1
Velocity is decreasing from a positive constant to some negative number with zero at t1
Distance from the original point increases from zero to t1 and then it decreases afterwards.
The object is moving in a positive direction at first and after time t1 it moves in a negative direction.
What does this graph indicate about the motion of an object ?
- a)object is moving in negative direction with positive acceleration
- b)object is moving in negative direction with negative acceleration
- c)object is moving in positive direction with negative acceleration
- d)object is moving in positive direction with positive acceleration
Correct answer is option 'C'. Can you explain this answer?
What does this graph indicate about the motion of an object ?
a)
object is moving in negative direction with positive acceleration
b)
object is moving in negative direction with negative acceleration
c)
object is moving in positive direction with negative acceleration
d)
object is moving in positive direction with positive acceleration
|
Raghav Bansal answered |
Velocity is decreasing so, acceleration is negative. Velocity is positive so direction is positive.
Hence option (C) is correct.
The position x of a particle varies with time (t) as x = 3t2 – 2t3 .The acceleration of the particle will be zero at time- a)1
- b)3/2
- c)0
- d)1/2
Correct answer is option 'D'. Can you explain this answer?
The position x of a particle varies with time (t) as x = 3t2 – 2t3 .The acceleration of the particle will be zero at time
a)
1
b)
3/2
c)
0
d)
1/2
|
Lavanya Menon answered |
We know that acceleration is the second derivative of change in displacement wrt time, hence we get a = 6 - 12t
When a = 0 = 6 -12t ;
We get t = ½
When a = 0 = 6 -12t ;
We get t = ½
The displacement of a particle along x-axis is given by x = 4 + 6t + 5t2. Its acceleration at t = 2s- a)2 m/s2
- b)10 m/s2
- c)1000 m/s2
- d)100 m/s2
Correct answer is option 'B'. Can you explain this answer?
The displacement of a particle along x-axis is given by x = 4 + 6t + 5t2. Its acceleration at t = 2s
a)
2 m/s2
b)
10 m/s2
c)
1000 m/s2
d)
100 m/s2
|
Mira Sharma answered |
usingdifferentiation
x = 18t + 5t^2
dx/dt = 18 + 10t
v = 18 + 10t
at t = 2s
v = 18 + 10x2
= 18 + 20 =38m/s
at t = 3s
v = 18 + 10x3
= 18 + 20 = 48m/s
vav= 38+48/2= 43m/s
dv/dt = 10
a = 10
a =10m/s^2
On acceleration-time graph the area under the curve equals the:- a)Meaningless
- b)Change in velocity
- c)Displacement
- d)Change in acceleration
Correct answer is option 'B'. Can you explain this answer?
On acceleration-time graph the area under the curve equals the:
a)
Meaningless
b)
Change in velocity
c)
Displacement
d)
Change in acceleration
|
|
Nandini Patel answered |
The area under an acceleration graph represents the change in velocity. In other words, the area under the acceleration graph for a certain time interval is equal to the change in velocity during that time interval.
Chapter doubts & questions for Acceleration - Physics for EmSAT Achieve 2025 is part of EmSAT Achieve exam preparation. The chapters have been prepared according to the EmSAT Achieve exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for EmSAT Achieve 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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