Q1. A particle is moving in a circular path of radius r. The displacement after half a
circle would be:
(a) Zero (b) π r (c) 2 r (d) 2π r
Q2. The numerical ratio of displacement to distance for a moving object is
(a) always less than 1 (b) always equal to 1
(c) always more than 1 (d) equal or less than 1
Q3. i)Classify the given quantities as Scalar or Vector and Write their S.I unit :
Distance, displacement, speed , velocity , acceleration.
ii) Write the formulas of the following quantities : speed, velocity, average speed ,
average velocity, acceleration, and 3 equation of motion.
Q4. A ball is thrown upwards and it goes to the height 100 m and comes down.
i) What is the net displacement?
ii) What is the net distance?
Q5. i) What can you say about motion of an object whose distance time graph is a straight line parallel to the time axis?
ii) Which quantity is measured by the area occupied under velocity- time graph?
Q6. Draw a velocity –time graph of a stone thrown vertically upwards and then coming downwards after attaining the maximum height.
Q7. Kiara hired a taxi and noted the initial reading of its odometer as 24803 km. After the journey, she noted the final reading of the odometer as 25118 km. If the journey took 4 hours 40 minutes, Calculate the average speed of the taxi during the entire journey.
Q8. The velocity-time graph (Fig. ) shows the motion of a cyclist. Find (i) its acceleration (ii) the distance covered by the cyclist in 15 second
Q9. A body travels 50 km at an average speed of 60km/h and returns at an average speed of 80km/h . Find average speed of the whole journey.
Q10. A marble rolling on a smooth floor has an initial velocity of 0.4m/s. If the floor offers a retardation of 0.02m/s2, Calculate the time it will take to come to rest.
Q11. Consider a car being driven along a straight road. The velocity of the car in m/s at different instants of time is shown in the following table. Draw velocity –time graph for motion of the car. Obtain the value of acceleration of the car from the graph.
Time(s) 0 10 20 30 40 50
Velocity(m/s) 0 5 10 15 20 25
Q12. The velocity time graph of two bodies A and B traveling along the +x direction are given in the figure
(a) Are the bodies moving with uniform acceleration?
(b) Which body is moving with greater acceleration A or B? Give reason
WHO WILL GIVE CORRECT ANSWER WILL BE MARKED AS BRAINLIEST!!!!!!!!!!!!
Answers
Explanation:
The answer will be 2r.
After the particle covers half path of the circle, the displacement can be easily calculated by measuring the distance between initial and final points. Displacement after half circle = AB=OA + OB Hence, the displacement after half circle is 2r.
Answer:
HEY THERE!!
1. Displacement at half path is the shortest distance between the initial and final point.
So, it's equal to twice the radius of the circle =2r
2. Magnitude of displacement is the staright line distance between the initial point and the final point of the path of the moving body.
Dispalcement could be negative, positive or zero. but distance is always positive or zero when there is no motion.
So, the ratio of displacement to distance for a moving object is equal to less than 1.
3.i) VECTOR SI UNIT SCALAR SI UNIT
Velocity m/s Distance m
Acceleration m/s2 Speed m/s
Displacement m
ii) Speed = distance/time
velocity=displacement/time
average speed=distance/time
average velocity=initial position-final position/time interval (xf-xi/tf-ti)
acceleration=v-u/t
5.i.) When the slope of distance-time graph is a straight line parallel to time axis, the object is at the same position as the time passes. That means the object is at rest.
ii.)displacement
7.)Given:
Initial reading = 24803 km
Final reading = 25118 km
Solution:
The speedometer reading at the start of the journey and at the end of the journey was noted by Kiara and the total time was also noted, the average speed is given by,
Total time = 4 hr and 40 min
avg. speed = total distance/total time
Total distance = Final reading – Initial reading
Total distance = 25118 – 24803
Total distance = 315 km
Total time = 4 hr 40 min
total time=4hr+(40/60)hr
Total time = 4.67 hr
avg speed =315/4.67
Average speed = 67.45 km/hr
8.)According to figure,
(i). The acceleration will be zero because the velocity is constant.
(II). The velocity is 20 m/s.
(III). The distance is equal to the product of the time and velocity.
The distance is
D=vt
=20*10
=200m
Hence, The acceleration is zero ,the velocity is 20 m/s and the distance covered by the cyclist is 200 m after 15 sec.
9.)let the time be t1 to travel to the point and time to return be t1. Then,
distance travelled in first half = 50 km
distance travelled in second half = 50 km
First Half:
50km/t1 = 60km/hr
OR t1 = 50/60 = 5/6hr
Second Half:
50km/t2 = 80km/hr
OR t2 = 50/80 = 5/8hr
Overall distance:
50+50 = 100km
Overall time:
t1+t2 = 5/6 + 5/8 = 35/24
Average Speed :
(100km*24)/35 = 480/7 = 68.57 km/hr. (approx)
10)The initial velocity of marble is 0.4m/s.
And the retardation is 0.2m/s²,or a=-0.2m/s²
Now by first equation of motion
v=u+at
where u is initial velocity,v is final velocity a is acceleration and t is time.
Putting the values in the equation,we get
0=0.4+(-0.2)t. (final velocity is 0 because body comes to rest.)
or,0.2t=0.4
or,t=0.4/0.2=2s
12.) a)yes
b)Slope of A > Slope of B
Slope of A = x/t = v
So Change in velocity is also great along A.
So A has more acceleration.
HOPE IT HELPS!!