In the previous section, we derived the three equations of motion. They are:
■ First equation of motion: v = u + at
■ Second equation of motion: s = ut + 1⁄2 at2
■ Third equation of motion: v2 = u2 + 2as
In this section we will see some solved examples.
Solved example 1.8
A train starting from rest attains a velocity of 72 kmph in 5 minutes. Assuming that the acceleration is uniform, find (i) the acceleration and (ii) the distance travelled by the train for attaining this velocity.
Solution:
Part (i):
1. The train starts from rest. So u = 0
2. The final velocity attained in 5 minutes is 72 kmph. So v = 72 kmph.
3. And t = 5 minutes
• Note the units: Speed is given in kmph and time is given in minutes
• But 'minutes' is not generally used when we specify velocity. 'Seconds' or 'hours' are used
♦ In kmph we use hours
♦ In m s-1 we use seconds
• So we have to change the '5 minutes' to either hours or seconds. In this problem, it is convenient to use seconds. So we get:
t = 5 minutes = 5 × 60 = 300 Seconds
4. Since time is in seconds we must convert velocity into m s-1. We get:
72 kmph = (72 × 1000)⁄(1 × 3600) = 20 m s-1
5. So we have u, v and t. We have to find the acceleration a
6. Out of the three equations of motion, the one which connects the four quantities in (5), is the first equation. So we will use it.
7. v = u + at ⇒ a = (v-u)⁄t ⇒ a = (20-0)⁄300 = 1⁄15 m s-2.
Part (ii):
1. We have u, v, t and a. We have to find s.
2. Out of the three equations of motion, there is none which connects the five quantities in (1). But we do not need all the five quantities to be connected. The second equation will help us to get the result. So we will use it.
3. s = ut + 1⁄2 at2 ⇒ s = 0 × 300 + 1⁄2 ×(1⁄15)×3002 ⇒ s = 3000 m = 3 Km
■ So the train was able to increase it's velocity from zero to 20 m s-1 within a distance of 3 Km because, it was subjected to an acceleration of 1⁄15 m s-2 . And the train took 300 seconds (5 minutes) to attain the velocity of 20 m s-1.
Solved example 1.9
A car accelerates uniformly from 18 kmph to 36 kmph in 5 s. Calculate (i) the acceleration and (ii) the distance covered by the car in that time.
Solution:
Part (i):
1. u = 18 kmph = (18 × 1000)⁄(1 × 3600) = 5 m s-1
2. v = 36 kmph = (36 × 1000)⁄(1 × 3600) = 10 m s-1
3. t = 5 s
4. We have to find acceleration a. Out of the three equations of motion, the one which connects the above three quantities and a is, the first equation. So we will use it.
5. v = u + at ⇒ a = (v-u)⁄t ⇒ a = (10-5)⁄5 = 5⁄5 = 1 m s-2.
Part (ii):
1. We have u, v, t and a. We have to find s.
2. Out of the three equations of motion, there is none which connects the five quantities in (1). But we do not need all the five quantities to be connected. The second equation will help us to get the result. So we will use it.
3. s = ut + 1⁄2 at2 ⇒ s = 5 × 5 + 1⁄2 × 1 × 52 ⇒ s = 37.5 m
■ So the car was able to increase it's velocity from 5 m s-1 to 10 m s-1 within a distance of 37.5 m because, it was subjected to an acceleration of 1 m s-2 . And the car took 5 seconds to attain the velocity of 10 m s-1.
Solved example 1.10
The brakes applied to a car produce an acceleration of 6 m s-2 in the opposite direction to the motion. If the car takes 2 s to stop after the application of brakes, calculate the distance it travels during this time.
Solution:
1. a = -6 m s-2. The negative sign is applied because it is deceleration. It is opposing the motion.
2. t = 2 s
3. We have to find the distance s.
• Consider the three equations of motion. Only the second one connects s, a and t.
• So we will have to use it. But to use it, we need to find the initial velocity u. It is the velocity at which the car was travelling just before brakes began to slow it down.
• So we will use the first equation to find that u:
4. v = u + at ⇒ u = v - at ⇒ u = 0 - (-6) × 2 ⇒ u = 12 m s-1. (∵ final velocity is zero as the car comes to a stop)
5. Now we can use the second equation:
s = ut + 1⁄2 at2 ⇒ s = 12 × 2 + 1⁄2 × (-6) × 22 ⇒ s = 24 + (-12) = 12 m
6. Note that, once we find u, we can use the third equation also:
v2 = u2 + 2as ⇒ 02 = 122 + 2 × (-6) × s ⇒ 0 = 144 -12s ⇒ 12s = 144 ⇒ s = 12 m
■ So the car was able to decrease it's velocity from 12 m s-1 to 0 m s-1 because, it was subjected to a deceleration of (-6) m s-2 . And the car took 2 seconds to come to a stop. Also note that the car travelled 12 m during this 2 seconds. This is why drivers are cautioned to maintain some distance between vehicles while travelling on the road.
Solved example 1.11
A bus starting from rest moves with a uniform acceleration of 0.1 m s-2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.
Solution:
Part (i):
1. u = 0, a = 0.1 m s-2, t = 120 s
2. We have to find the final speed v.
3. We can use the first equation of motion. It connects all the above four quantities.
4. v = u + at ⇒v = 0 + 0.1 × 120 = 12 m s-1.
Part (ii):
1. We have to find s. We can use the second equation:
s = ut + 1⁄2 at2 ⇒ s = 0 × 120 + 1⁄2 × 0.1 × 1202 ⇒ s = 0 + 120 × 6 = 720 m
Solved example 1.12
A train is travelling at a speed of 90 kmph. Brakes are applied so as to produce a uniform acceleration of (-0.5) m s-2. Find how far the train will go before it is brought to rest.
Solution:
1. a = -0.5 m s-2. The negative sign is applied because it is deceleration. It is opposing the motion.
2. t = 2 s
3. u = 90 kmph = (90 × 1000)⁄(1 × 3600) = 25 m s-1
4. We have to find the distance s. We can use the third equation of motion.
5. v2 = u2 + 2as ⇒ 02 = 252 + 2 × (-0.5) × s ⇒ 0 = 625 - s ⇒ s = 625 m
Solved example 1.13
A trolley, while going down an inclined plane, has an acceleration of 2 cm s-2. What will be its velocity 3 s after the start?
Solution:
1. u = 0, a = 2 cm s-2, t = 3 s
2. We have to find the final speed v.
3. We can use the first equation of motion. It connects all the above four quantities.
4. But before that, we have to convert the acceleration from cm s-2 to m s-2
We have: 2 cm = 2⁄100 m = 0.02 m. So 2 cm s-2 = 0.02 m s-2.
5. v = u + at ⇒v = 0 + 0.02 × 3 = 0.06 m s-1.
Solved example 1.14
A racing car has a uniform acceleration of 4 m s-2. What distance will it cover in 10 s after start?
Solution:
1. u = 0, a = 4 m s-2, t = 10 s
2. We have to find the distance s. We can use the second equation of motion:
3. s = ut + 1⁄2 at2 ⇒s = 0 + 1⁄2 × 4 × 102 ⇒ s = 200 m
Solved example 1.15
A stone is thrown in a vertically upward direction with a velocity of 5 m s-1. If the acceleration of the stone during its motion is 10 m s-2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Solution:
1. u = 5 m s-1, a = -10 m s-2.
2. We have to find s and t. We can not use the second equation s = ut + 1⁄2 at2 because it has two unknowns s and t. Let us try the third equation:
3. v2 = u2 + 2as. In this also there seems to be two unknowns v and s. But in reality, v is not an unknown. Let us analyse:
• The maximum height reached by the stone is the s that we want.
• When the stone is thrown upwards, it's velocity continuously decreases because of the deceleration of 10 ms-2.
• The velocity continuously decreases and finally becomes zero. So v = 0. After that, it falls downwards.
4. So we can use the third equation:
v2 = u2 + 2as ⇒ 02 = 52 + 2 × (-10) × s ⇒ 0 = 25 - 20s ⇒ s = 1.25 m
5. To find the time, we can use the first equation:
v = u + at ⇒0 = 5 + (-10) × t ⇒0 = 5 -10t ⇒5 = 10t ⇒t = 0.5 s
So we have completed the discussion on the 'Motion along a straight line'. In the next section, we will see 'Motion along a circular path'.
■ First equation of motion: v = u + at
■ Second equation of motion: s = ut + 1⁄2 at2
■ Third equation of motion: v2 = u2 + 2as
In this section we will see some solved examples.
Solved example 1.8
A train starting from rest attains a velocity of 72 kmph in 5 minutes. Assuming that the acceleration is uniform, find (i) the acceleration and (ii) the distance travelled by the train for attaining this velocity.
Solution:
Part (i):
1. The train starts from rest. So u = 0
2. The final velocity attained in 5 minutes is 72 kmph. So v = 72 kmph.
3. And t = 5 minutes
• Note the units: Speed is given in kmph and time is given in minutes
• But 'minutes' is not generally used when we specify velocity. 'Seconds' or 'hours' are used
♦ In kmph we use hours
♦ In m s-1 we use seconds
• So we have to change the '5 minutes' to either hours or seconds. In this problem, it is convenient to use seconds. So we get:
t = 5 minutes = 5 × 60 = 300 Seconds
4. Since time is in seconds we must convert velocity into m s-1. We get:
72 kmph = (72 × 1000)⁄(1 × 3600) = 20 m s-1
5. So we have u, v and t. We have to find the acceleration a
6. Out of the three equations of motion, the one which connects the four quantities in (5), is the first equation. So we will use it.
7. v = u + at ⇒ a = (v-u)⁄t ⇒ a = (20-0)⁄300 = 1⁄15 m s-2.
Part (ii):
1. We have u, v, t and a. We have to find s.
2. Out of the three equations of motion, there is none which connects the five quantities in (1). But we do not need all the five quantities to be connected. The second equation will help us to get the result. So we will use it.
3. s = ut + 1⁄2 at2 ⇒ s = 0 × 300 + 1⁄2 ×(1⁄15)×3002 ⇒ s = 3000 m = 3 Km
■ So the train was able to increase it's velocity from zero to 20 m s-1 within a distance of 3 Km because, it was subjected to an acceleration of 1⁄15 m s-2 . And the train took 300 seconds (5 minutes) to attain the velocity of 20 m s-1.
Solved example 1.9
A car accelerates uniformly from 18 kmph to 36 kmph in 5 s. Calculate (i) the acceleration and (ii) the distance covered by the car in that time.
Solution:
Part (i):
1. u = 18 kmph = (18 × 1000)⁄(1 × 3600) = 5 m s-1
2. v = 36 kmph = (36 × 1000)⁄(1 × 3600) = 10 m s-1
3. t = 5 s
4. We have to find acceleration a. Out of the three equations of motion, the one which connects the above three quantities and a is, the first equation. So we will use it.
5. v = u + at ⇒ a = (v-u)⁄t ⇒ a = (10-5)⁄5 = 5⁄5 = 1 m s-2.
Part (ii):
1. We have u, v, t and a. We have to find s.
2. Out of the three equations of motion, there is none which connects the five quantities in (1). But we do not need all the five quantities to be connected. The second equation will help us to get the result. So we will use it.
3. s = ut + 1⁄2 at2 ⇒ s = 5 × 5 + 1⁄2 × 1 × 52 ⇒ s = 37.5 m
■ So the car was able to increase it's velocity from 5 m s-1 to 10 m s-1 within a distance of 37.5 m because, it was subjected to an acceleration of 1 m s-2 . And the car took 5 seconds to attain the velocity of 10 m s-1.
Solved example 1.10
The brakes applied to a car produce an acceleration of 6 m s-2 in the opposite direction to the motion. If the car takes 2 s to stop after the application of brakes, calculate the distance it travels during this time.
Solution:
1. a = -6 m s-2. The negative sign is applied because it is deceleration. It is opposing the motion.
2. t = 2 s
3. We have to find the distance s.
• Consider the three equations of motion. Only the second one connects s, a and t.
• So we will have to use it. But to use it, we need to find the initial velocity u. It is the velocity at which the car was travelling just before brakes began to slow it down.
• So we will use the first equation to find that u:
4. v = u + at ⇒ u = v - at ⇒ u = 0 - (-6) × 2 ⇒ u = 12 m s-1. (∵ final velocity is zero as the car comes to a stop)
5. Now we can use the second equation:
s = ut + 1⁄2 at2 ⇒ s = 12 × 2 + 1⁄2 × (-6) × 22 ⇒ s = 24 + (-12) = 12 m
6. Note that, once we find u, we can use the third equation also:
v2 = u2 + 2as ⇒ 02 = 122 + 2 × (-6) × s ⇒ 0 = 144 -12s ⇒ 12s = 144 ⇒ s = 12 m
■ So the car was able to decrease it's velocity from 12 m s-1 to 0 m s-1 because, it was subjected to a deceleration of (-6) m s-2 . And the car took 2 seconds to come to a stop. Also note that the car travelled 12 m during this 2 seconds. This is why drivers are cautioned to maintain some distance between vehicles while travelling on the road.
Solved example 1.11
A bus starting from rest moves with a uniform acceleration of 0.1 m s-2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.
Solution:
Part (i):
1. u = 0, a = 0.1 m s-2, t = 120 s
2. We have to find the final speed v.
3. We can use the first equation of motion. It connects all the above four quantities.
4. v = u + at ⇒v = 0 + 0.1 × 120 = 12 m s-1.
Part (ii):
1. We have to find s. We can use the second equation:
s = ut + 1⁄2 at2 ⇒ s = 0 × 120 + 1⁄2 × 0.1 × 1202 ⇒ s = 0 + 120 × 6 = 720 m
Solved example 1.12
A train is travelling at a speed of 90 kmph. Brakes are applied so as to produce a uniform acceleration of (-0.5) m s-2. Find how far the train will go before it is brought to rest.
Solution:
1. a = -0.5 m s-2. The negative sign is applied because it is deceleration. It is opposing the motion.
2. t = 2 s
3. u = 90 kmph = (90 × 1000)⁄(1 × 3600) = 25 m s-1
4. We have to find the distance s. We can use the third equation of motion.
5. v2 = u2 + 2as ⇒ 02 = 252 + 2 × (-0.5) × s ⇒ 0 = 625 - s ⇒ s = 625 m
Solved example 1.13
A trolley, while going down an inclined plane, has an acceleration of 2 cm s-2. What will be its velocity 3 s after the start?
Solution:
1. u = 0, a = 2 cm s-2, t = 3 s
2. We have to find the final speed v.
3. We can use the first equation of motion. It connects all the above four quantities.
4. But before that, we have to convert the acceleration from cm s-2 to m s-2
We have: 2 cm = 2⁄100 m = 0.02 m. So 2 cm s-2 = 0.02 m s-2.
5. v = u + at ⇒v = 0 + 0.02 × 3 = 0.06 m s-1.
Solved example 1.14
A racing car has a uniform acceleration of 4 m s-2. What distance will it cover in 10 s after start?
Solution:
1. u = 0, a = 4 m s-2, t = 10 s
2. We have to find the distance s. We can use the second equation of motion:
3. s = ut + 1⁄2 at2 ⇒s = 0 + 1⁄2 × 4 × 102 ⇒ s = 200 m
Solved example 1.15
A stone is thrown in a vertically upward direction with a velocity of 5 m s-1. If the acceleration of the stone during its motion is 10 m s-2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Solution:
1. u = 5 m s-1, a = -10 m s-2.
2. We have to find s and t. We can not use the second equation s = ut + 1⁄2 at2 because it has two unknowns s and t. Let us try the third equation:
3. v2 = u2 + 2as. In this also there seems to be two unknowns v and s. But in reality, v is not an unknown. Let us analyse:
• The maximum height reached by the stone is the s that we want.
• When the stone is thrown upwards, it's velocity continuously decreases because of the deceleration of 10 ms-2.
• The velocity continuously decreases and finally becomes zero. So v = 0. After that, it falls downwards.
4. So we can use the third equation:
v2 = u2 + 2as ⇒ 02 = 52 + 2 × (-10) × s ⇒ 0 = 25 - 20s ⇒ s = 1.25 m
5. To find the time, we can use the first equation:
v = u + at ⇒0 = 5 + (-10) × t ⇒0 = 5 -10t ⇒5 = 10t ⇒t = 0.5 s
So we have completed the discussion on the 'Motion along a straight line'. In the next section, we will see 'Motion along a circular path'.
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