- An air force jet flying with a speed of 335m/s went past an anti aircraft gun. How far is the aircraft 5secs later when the gun was fired.
A. 838m. B. 3350m. C. 670m. D. 1675m. E. 67m
Solution
V = 335m/s,. Time = 5secs,
Formula, V = distance/time ( since we’re asked to find distance)
Distance = velocity×time; 335×5
= 1675m - A train has an initial velocity of 44m/s and an acceleration of -4m/s². It’s velocity after 10 sec is
A. 2m/s. B. 4m/s. C. 8m/s. D. 12m/s. E. 16m/s
Solution
U(initial velocity) = 44m/s, a = -4m/s², t = 10secs
Formula, a = (v-u)/t
-4 = (v-44)/10 ;
-4×10 = v-44
-40 = v-44
-40+44 = v
V = 4m/s - If a car starts from rest and moves with a uniform acceleration of 10m/s² for ten seconds the distance, it covers in the last one second of motion is
A. 95m. B. 100m. C. 500m. D. 905m. E. 1000m
Solution
a,(acceleration) = 10m/s², t=10secs, last one second of motion = subtract 10-9 seconds
Starts from rest, u = 0
Distance for ten seconds
Find distance, S=ut+½at²
S = 0×10+½×10×10²
S = 0+½×10×100
S1 = 500m
Find distance, S=ut+½at²
S = 0×10+½×10×9²
S2 = 0+½×10×81
= 405m
S =( distance for ten seconds)-(distance for nine second)
S = 500-405
= 95m. (A)
4.The figure below (Fig. 1.13) shows the velocity-time graph of a car which starts from rest and is accelerated uniformly at the rate of 3m/s² for 5 seconds. It attains a velocity which is maintained for 1 minute. The car is then brought to rest by a uniform retardation after another 3 seconds. Calculate the total distance covered.
A. 900meters B. 1920meters C. 1020meters D. 960meters E. 860meters
Solution
A = 3m/s², t = 5secs
V = u+ at
=0 + 3×5
V = 15m/s²
S = ½(a+b) ×h
½ (60+68) ×15 = ½ (128) ×15
960m
- The graph below (Fig. 1.14) describes the motion of a particle. The acceleration of the particle during the motion is
A. 0.00ms–². B. 0.25ms–² C. 4.00ms–². D. 8.00ms–². E. 10.00ms–²
Solution
. The acceleration of the particle during the motion = change in y velocity (ms-2) / change in x time(t seconds) a = (10-2)/(2-0); = 8/2
4.00ms–³
- What is the average velocity of the sprinter whose velocity-time graph is shown in the figure below?
A. 85.0ms-1 B. 17.0ms-2 C. 8.0ms-2 D. 1.7ms-2
Solution
Velocity time graph for the object in the graph at acceleration is = 10-2 ⇛ 8ms-2
- The diagram below (Fig. 1.17) shows a velocity-time graph representing the motion of a car. Find the total distance covered during the acceleration and retardation periods of the motion.
A. 75m B. 150m C. 300m D. 375m
Solution
Change in time = T2-T1 ⇛ 10-0 ⇛ 10m/s
For acceleration, a = (v-u)/t; = (10-0)/10
a = 1m/s
distances 1, s = ut+1/2 at2
0×10+1/2 ×1v102 = 50m
For deceleration, a =(u-v)/t
Change in time = 45-40 ⇛ 5secs
Deceleration, a = (10-0)/5
2m/s
Distance 2, s = ut + ½ ×2×52
25m
Total distance = s1 + s2
75m
- A body falls freely under gravity ( g = 9.8m/s2) from a height of 40m on to the top of a platform 0.8m above the ground. It’s velocity on reaching the platform is
A. 784m/s B. 80m/s C. 78.4m/s D. 39.2m/s E. 27.7m/s
Solution
G = 9.8m/s2, height = 40m, total height = 40-0.8; = 39.2m
V2 = U2 + 2GS
V2 = 0 + 2×9.8×39.2
= 27.7m/s
9. The figure below (Fig. 1.13) shows the velocity-time graph of a car which starts from rest and is accelerated uniformly at the rate of 3m/s2 for 5 seconds. It attains a velocity which is maintained for 1 minute. The car is then brought to rest by a uniform retardation after another 3 seconds. Calculate the total distance covered.
A. 900meters B. 1920meters C. 1020meters D. 960meters E. 860meters
Solution
A = 3m/s2, t = 5secs
V = u+ at
=0 + 3×5
V = 15m/s2
S = ½(a+b) ×h
½ (60+68) ×15 = ½ (128) ×15
960m
- The graph below (Fig. 1.14) describes the motion of a particle. The acceleration of the particle during the motion is
A. 0.00ms-². B. 0.25ms-². C. 4.00ms-2² D. 8.00ms-² E. 10.00ms-²
Solution
. The acceleration of the particle during the motion = change in y velocity (ms-2) / change in x time(t seconds) a = (10-2)/(2-0); = 8/2
4.00ms-²b
- What is the average velocity of the sprinter whose velocity-time graph is shown in the figure above?
A. 85.0ms-1 B. 17.0ms-2 C. 8.0ms-2 D. 1.7ms-2
Solution
Velocity time graph for the object in the graph at acceleration is = 10-2 ⇛ 8ms-² - The diagram above (Fig. 1.17) shows a velocity-time graph representing the motion of a car. Find the total distance covered during the acceleration and retardation periods of the motion.
A. 75m B. 150m C. 300m D. 375m
Solution
Change in time = T2-T1 ⇛ 10-0 ⇛ 10m/s
For acceleration, a = (v-u)/t; = (10-0)/10
a = 1m/s
distances 1, s = ut+1/2 at2
0×10+1/2 ×1v102 = 50m
For deceleration, a =(u-v)/t
Change in time = 45-40 ⇛ 5secs
Deceleration, a = (10-0)/5
2m/s
Distance 2, s = ut + ½ ×2×52
25m
Total distance = s1 + s2
75m
- A body falls freely under gravity ( g = 9.8m/s2) from a height of 40m on to the top of a platform 0.8m above the ground. It’s velocity on reaching the platform is
A. 784m/s B. 80m/s C. 78.4m/s D. 39.2m/s E. 27.7m/s
Solution
G = 9.8m/s2, height = 40m, total height = 40-0.8; = 39.2m
V2 = U2 + 2GS
V2 = 02 + 2×9.8×39.2
= 27.7m/s
- A body falls freely under gravity ( g = 9.8m/s2) from a height of 40m on to the top of a platform 0.8m above the ground. It’s velocity on reaching the platform is
A. 784m/s B. 80m/s C. 78.4m/s D. 39.2m/s E. 27.7m/s
Solution
G = 9.8m/s2, height = 40m, total height = 40-0.8; = 39.2m
V2 = U2 + 2GS
V2 = 02 + 2×9.8×39.2
= 27.7m/s - The diagram below (Fig 1.18) shows the velocity-time graph of a vehicle. Its acceleration and retardation respectively are.
A. 8.0ms-², 4.0 ms–². B.. 4.0 ms-², 8.0 ms-² C. 4.0 ms-², 2.0 ms-². D. 2.0 ms-² , 4.0 ms-²
Solution
At acceleration the velocity-time graph = change in velocity (m/s)/ change in time ( seconds)
(V2-V1)/(T1-T2) = (80-0)/(20-0)
4m/s²
Retardation of the velocity-time graph = change in velocity (m/s)/ change in time ( seconds)
(80-0)/(90-50) = 2m/s²
Its acceleration and retardation respectively are 4.0 ms-², 2.0 ms-²
- A mango fruit drops from a branch 10m above the ground, just before hitting the ground its velocity is
A. 10-1 B. 10/-1 C. 100-1 D. 5-1
Solution
The answer is 10-1 - A palm fruit dropped to the ground from the top of a tree 45m tall. How long does it take to reach the ground? [g = 10ms-2]A. A. 9s B. 4.5s C. 6s D. 7.5s E. 3s
Solution
H = ut + 1/2gt2
45 =0×t + ½ ×10×t2
t2 = 9
t = 3 - A small metal ball is thrown vertically upwards from the top of a tower with an initial velocity of 20ms-1, if the ball took a total of 6 seconds to reach the ground level, determine the height of the tower. (g = 10ms-2)
A. 60m B. 60m D. 100m D. 20m
Solution
Time = 6 seconds, u = 20ms-2
V2 = U2 + 2gh
02 = 202 + 2×10×h
H = 400/20
H = 20m - In free fall, a body of mass 1k drops from a height of 125m from rest in 5 seconds. How long will it take another body of mass 2kg to fall from rest, from the same height? [g – 10ms-2]A. 5s B. 10s c. 12s D. 15s
Solution
Since it is a free fall, using the law of gravity the answer will still be 5s
Answer = 5s - Two particles X and Y starting from rest cover the same distance. The acceleration of X is twice that of Y. the ratio of the time taken by X to that taken by Y is?
A. ½ n B. 2 C. 1/ D. E. 4
Solution
The acceleration of X is twice that of Y = 1/ - A bullet fired vertically upward from a gun held 2.0m above the ground reaches its maximum height in 4.0s. Calculate its initial velocity. (g = 10ms-2)
A. 10ms-1 B. 8 ms-1 C. 20 ms-1 D. 40 ms-1
Solution
Since it’s vertically upward, U = V + at
U = 0 + 10×4
U = 40 ms-1 - What is the acceleration between two points on a velocity-time graph which has coordinates (10s, 15 ms-1 ) and (20s, 35 ms-1 )?
A. 1.75 ms-2 B. 3.50 ms-2 C. 1.00 ms-2 D. 2.00 ms-2
Solution
Acceleration = change in velocity (m/s)/ change in time ( seconds)
(35-15)/ (20-10) = 20/10
2 ms-2 - A car accelerates uniformly from rest at 4 ms-2. How far will it travel in fifth complete second?
A. 100m B. 50m C. 32m D. 18m
Solution
travel in fifth complete second = T5-T4
S = ut + ½ at2
S = 0×t + ½ ×4×(52-42)
S = ½ × 4×(25-16)
S = 2×9
S = 18m
