94. A boy walks to his school at a distance of 6km with constant speed of 2.5kmph and walks back with a
constant speed of 4kmph. His average speed for round trip expressed in kmph is
1) 24/13 2) 40/13 3) 3 4) 1/2
95. If u 2i 2j 3k and the final velocity is v 2i 4j 5k and it is covered in a time of 10 sec, find the
acceleration vector.
1)
10
3i 2j 2k
2)
10
3i j 2k
3)
10
3i 2j 2k
4)
5
j k
96. A train moving at a constant velocity of 54 km/hr moves east wards for 30 minutes, then due north with the
same speed for 40 minutes. What is the average velocity of the train during this run? (in km/hr)
1) 30 2) 35 3) 38.6 4) 49.3
97. A car moving with constant acceleration covers the distance between two points 180 m apart in 6 sec. Its
speed as it passes the second point is 45 m/s. What is its acceleration and its speed at the first point
1) -5 2 m/ s ; 15 m/s 2) -15 2 m/ s ; 5 m/s 3) -5 2 m/ s ; -15 m/s 4) 5 2 m/ s ; 15 m/s
98. A body moving with a uniform acceleration had velocities of 20 m/s and 30 m/s when passing the
points P and Q of its path. Find the velocity midway between P and Q ( in m/s)
1) 450 2) 550 3) 650 4) none
99. A particle starts moving from rest with uniform acceleration. It travels a distance x in the first 2 sec and a
distance y in the next 2 sec. Then
1) y = x 2) y = 2x 3) y = 3x 4) y = 4x
100. A bullet fired into a fixed target loses half of its velocity in penetrating 15 cm. How much further it will penetrate before
coming to rest?
1) 5 cm 2) 15 cm 3) 7.5 cm 4) 10 cm
101. For a body travelling with uniform acceleration, its final velocity is v 180 7x , where x is the distance
travelled by the body. Then the acceleration is
1) -8 2 m/ s 2) -3.5 2 m/ s 3) -7 2 m/ s 4) 180 2 m/ s
102. A motorist drives north for 35.0 minutes at 85.0 km/h and then stops for 15.0 minutes. He next continues
north, travelling 130 km in 2.00 hours. a) What is his total displacement
1)85 km 2) 179.6 km 3)20 km 4) 140 km
103. The position x of a particle varies with time t as x = at2
– bt3
. The acceleration of the particle will be zero at
time t equal to
1)
a
b
2)
2a
3b 3)
a
3b 4) Zero
Answers
Explanation:
94)-
Constant speed = 2.5km/h
Distance = 6km
Time taken = 6/2.5
= 2.4 hours
Going from school to home-
Constant speed = 4km/h
Distance = 6km
Time taken = 6/4
= 1.5 hours
Total time is taken in traveling = 3.9 hours
Total distance = 12km
Average speed of round-trip = 12/3.9
=40/13 Ans:-(B)
96)distance covered due east= 54km/hr × 1/2 hr=27 km
distance covered due north=54km/hr × 4/6hr=36 km
displacement is the hypotenuse between starting and initial point.
=√27² + 36² = 45km
total time= 70 mins or 7/6 hrs
avg velocity=total displacement/time=45*6/7=38.4 km/hr
so ∴ avg velocity is 38.4 km/hr.
Ans:-C
97)
98)In usual notation,
v2 - u2 = 2as = 900 - 400 = 500
Thus, as = 250 --- (1)
At s/2 if speed is w:
w2 - u2 = 2a(s/2)
Which means,
w2 = u2 + as = 400 + 250
=650 Ans:-C
99) check in the attachment for solution the answer is (D)
Hello vihitha please ask the question again to solve the remaining questions because of the attachment problem if you ask again I will answer all the remaining questions
Answer:
Time for forward journey t
1
=
2.5
6
=
5
12
=2.4 hr
Time for backward journey t
2
=
4
6
= 1.5 hr
Average speed =
2.4+1.5
6+6
=
3.9
12
=
13
40
km/hr