Problems for Motion with Constant Acceleration in One Dimension

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Question 9.A truck accelerates from rest at the constant rate a for some time after which it decelerates at a constant rate of b to come to the rest.If the total time elapsed is t ,then find out the maximum velocity attains by the truck
a. (ab/a+b)t
c. (a2+b2/ab)t

. Let t1 and t2 be the the time for acceleration and deccleration.
Let v be the maximum velocity attained

v=at1 or t1=v/a
v=bt2 or t2=v/b

Now t=t1 + t2
or t=v/a + v/b

or v=abt/(a+b)
Hence (a) is correct

Question 10.Displacement(y) of the particle is given by
The velocity of the particle when acceleration is zero is given by
a. 5/2
b. 9/4
c. 13/6
d. 17/8

. given


Velocity is given


Now acceleratio is zero

Putting this value velocity equation
Hence (c) is correct

Question 11.Mark out the correct statement
a. Instantaneous Velocity vector is always in the direction of the motion
b. Instantaneous acceleration vector is always in the direction of the motion
c. Acceleration of the moving particle can change its direction without any change in direction of velocity
d. None of the above

. Take the case of uniform circular motion,Instantanous Velocity vector and acceleration vector at any point is tangent and radial to the circle.So it is not along the direction of the circle
Take the case the moving car in one direction.If the car accelerated,acceleration is along the direction of velocity.if car driver put a brake then it deacclerates without any change in the direction of the velocity

Hence (c) is correct

Matrix match type
Question 12.In a free fall motion from rest,Match column I to column II

column I
A) Graph between displacement and time
B) Graph between velocity and time
C) Graph between velocity and displacement
D) Graph between KE and displacement

column II
P) Parabola
Q) Straight line
C) Circle
D) No appropiate match given

. Equation of motion for a free fall from rest
x=(1/2)gt2.It is a parabola
v=gt it is a straight line
v2=2gx it is a parabola
KE=(1/2)mv2=mgx..... it is a straight line
So the correct answers are A -> P
B -> Q
C -> P
D -> Q

More Practice Questions
Question 1
A particle moves along the x axis. Its position id given by the equation x = 2 + 3t ? 4t2, with x in metres and t in seconds. Determine
(a) its position when it changes direction
(b) its velocity when it returns to the position it had at time t = 0.
a) x = 2 + 3t - 4t2
velocity (v) = dx/dt = 3-8t
Change in Direction will happen when velocity becomes zero and start becoming negative
So 3-8t=0
t= 3/8 sec

Position will be given as
x= 2+ 3(3/8) -4(3/8)2
x= 41/16 m
b) Position at t=0
x= 2 m

Now we need to find what all time x=2
2== 2 + 3t ? 4t2
t=0 or t =3/4
Now velocity at t=3/4
velocity (v) = dx/dt = 3-8t
= 3-8(3/4)=1 m/s
Question 2
A car moving with a speed of 40km/h can be stopped by applying the brakes after atleast 2 m.If the same car is moving with the speed 80K/h,what is the minimum stopping distance?
8 m

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