Motion In a Straight Line-Quiz-11

Physics Quiz-11

Dear Readers,

If an object changes its position with respect to its surroundings with time, then it is called in motion. It is a change in the position of an object over time. Motion in a straight line is nothing but linear motion. As the name suggests, it’s in a particular straight line, thus it can be said that it uses only one dimension.

Q1. A 2 m wide truck is moving with a uniform speed v_0=8 ms^(-1) along a straight horizontal road. A pedestrian starts to cross the road with a uniform speed v when the truck is 4 m away from him. The minimum value of  v so that he can cross the road safely is

•  2.62 ms^(-1)
•  4.6 ms^(-1)
•  3.57 ms^(-1)
• 1.414 ms^(-1)

Solution

Let the man start crossing the road at an angle θ with the roadside. For safe crossing, the condition is that the man must cross the road by the time the truck describes the distance (4+2 cot⁡θ)

So, (4+2 cot⁡θ)/8=(2/sin⁡θ)/v  or v=8/(2 sin⁡〖θ+cos⁡θ 〗 )

For minimum v,dv/dθ=0

or (-8(2 cos⁡〖θ-sin⁡θ 〗))/(2 sin⁡〖θ+cos⁡θ 〗 )^2 =0 or 2 cos⁡〖θ-sin⁡〖θ=0〗 〗

or tan⁡〖θ=2〗, so sin⁡〖θ=2/√5〗,cos⁡〖θ=1/√5〗

v_min=8/(2(2/√5)+1/√5)=8/√5=3.57 ms^(-1)

Q2.A person moves 30 m north and then 20 m towards east and finally 30√2 m in south-west direction. The displacement of the person from the origin will be

•  10 m along north
•  10 m along south
•  10 m along west
•  Zero

Solution

AB=30 m,BC=20 m,CD=30 √2  

DBCE is isosceles (Fig), so

BE=BC=20 m

AE=AB-BE=30-20=10 m

EC=20 √2  m

ED=CD-EC=30 √2-20√2=10√2 m

DADE is isosceles, so AD=AE=10 m

Q3. A body is projected upwards with a velocity u. It passes through a certain point above the ground after t_1. The time after which the body passes through the same point during the return journey is

•  (u/g-t_1^2 )
•  2(u/g-t_1 )
• 3(u^2/g-t_1 )
• 3(u^2/g^2 -t_1 )

Solution

Suppose v be the velocity attained by the body after time t_1. Then v=u-gt_1 (i)

Let the body reach the same point at time t_2. Now velocity will be downwards with same magnitude v, then –v=u-gt_2   …(ii)

(i)-(ii) ⇒2v=g(t_2-t_1)

or t_2-t_1=2v/g=2/g  (u-gt_1 )=2(u/g-t_1 )

Q4. The velocity acquired by a body moving with uniform acceleration is 30 ms^(-1) in 2 s and 60 ms^(-1) in 2 s. The initial velocity is

•  Zero
• 2 ms^(-1)
•  3 ms^(-1)
•  10 ms^(-1)

Solution

30=u+a×2,60=u+a×4

Solve to get u=0

Q5.The acceleration will be positive in

•  (I) and (III)
•  (I) and (IV)
•  (II) and (IV)
•  None of these

Solution

In (I), slope is negative and its magnitude is decreasing with time. It means slope is increasing numerically. So velocity is increasing towards right, and so acceleration is positive.

In (IV), slope is positive and its magnitude is increasing with time. So velocity is increasing towards right, and so acceleration is positive.

Q6.  

The angle between velocity and acceleration during retarded motion is

•  180°
•  40°
•  45°
•  0°

Solution

During retarded motion, acceleration and velocity are in opposite directions

Q7.

Two trains A and B, 100 m and 60 m long, are moving in opposite directions on parallel tracks. The velocity of the shorter train is 3 times that of the longer one. If the trains take 4 s to cross each other, the velocities of the trains are

• V_A=10 ms^(-1),V_B=30 ms^(-1)
• V_A=2.5 ms^(-1),V_B=7.5 ms^(-1)
•  V_A=20 ms^(-1),V_B=60 ms^(-1)
•  V_A=5 ms^(-1),V_B=15 ms^(-1)

Solution

3V_A=V_B,S_rel=n_rel t ⇒100+60=(V_A+V_B)×4

Solve to get V_A=10 ms^(-1) and V_B=30 ms^(-1)

Q8. A body covers one-third of the distance with a velocity v_1, the second one-third of the distance with a velocity v_2, and the remaining distance with a velocity v_3. The average velocity is

•  (v_1+v_2+v_3)/3
• (3v_1 v_2 v_3)/(v_1 v_2+v_2 v_3+v_3 v_1 )
• (v_1 v_2+v_2 v_3+v_3 v_1)/3
•  (v_1 v_2 v_3)/3

Solution

t_1=(S/3)/v_1 ,t_2=(S/3)/v_2 ,t_3  (S/3)/v_3 

v_av=S/(t_1+t_2+t_3 )=(3v_1 v_2 v_3)/(v_1 v_2+v_2 v_3+v_3 v_1 )

Q9.

A ball is thrown upwards with speed v from the top of a tower and it reaches the ground with speed 3v. What is the height of the tower?

• 
   v^2/g
• (2v^2)/g
•  (4v^2)/g
•  (8v^2)/g

Solution
According to the third equation of motion,
v^2-u^2=2as
Given v=3v,u=v and a=g
or (3v)^2-v^2=2gs or s=(4v^2)/g

Q10. Two cars A and B are travelling in the same direction with velocities V_A and V_B  (V_A>V_B). When the car A is at a distance s behind car B, the driver of the car A applies the brakes producing a uniform retardation a; there will be no collision when

•  s<(V_A-V_B )^2/2a
• s=(V_A-V_B )^2/2a
•  s≥(V_A-V_B )^2/2a
• s≤(V_A-V_B )^2/2a

Solution

For no collision, the speed of car A should be reduced to v_B before the cars meet, i.e., final relative velocity of car A with respect to car B is zero, i.e., V_r=0

Here u_r=initial relative velocity =V_A-V_B

Relative acceleration =a_r=-a-0=-a

Let relative displacement =s_r

Then using the equation, v_r^2=u_r^2+2a_r s_r

0^2=(V_A-V_B )^2-2as_r or s_r=(V_A-V_B )^2/2a

For no collision, s_r≤s i.e., (V_A-V_B )^2/2a≤s