Shortcut Methods and Tricks to Solve Numerical Problems on Motion in a Straight Line

JEE Main Level:

1. To find the distance traveled by a car in 10 seconds, use the formula:

$$s=ut+1/2a{t}^2$$

Where,

• u = initial velocity (in this case, u = 0 m/s since the car starts from rest)
• t = time (in this case, t = 10 s)
• a = acceleration (in this case, a = 2 m/s²)

By substituting the values, we get:

$$s=0\ast 10+1/2\ast 2\ast {10}^2$$ $$s=100m$$ Therefore, the car travels a distance of 100 meters in 10 seconds.

2. To find the average speed of a train, use the formula:

$$Speed=Distance/Time$$

Where,

• Distance = 200 km
• Time = 2 hours (which must be converted to seconds for consistency: 2 * 3600 = 7200 seconds)

By substituting the values, we get:

$$Speed=\frac{200000m}{7200s}$$ $$Speed=\frac{50}{3}m/s$$

Therefore, the average speed of the train is 50/3 m/s.

3. To find the maximum height reached by a ball thrown vertically upwards, use the formula:

$$Maximumheight=(u^2/2g)$$

Where,

• u = initial velocity (in this case, u = 20 m/s)
• g = acceleration due to gravity (in this case, g = 9.8 m/s²)

By substituting the values, we get:

$$Maximumheight=({20}^2/2\times 9.8)$$ $$Maximumheight=\frac{400}{19.6}$$ $$Maximumheight=20.45m$$

Therefore, the ball reaches a maximum height of approximately 20 meters.

4. To find the velocity with which a stone hits the ground, use the formula:

$$Finalvelocity=\surd (u^2+2gh)$$

Where,

• u = initial velocity (in this case, u = 0 m/s since the stone is dropped)
• g = acceleration due to gravity (in this case, g = 9.8 m/s²)
• h = height from which the stone is dropped (in this case, h = 100 m)

By substituting the values, we get:

$$Finalvelocity=\surd (0^2+2\times 9.8\times 100)$$ $$Finalvelocity=\surd (1960)$$ $$Finalvelocity=44.3m/s$$

Therefore, the stone hits the ground with a velocity of approximately 44.3 m/s.

5. To find the displacement of a particle moving with a constant acceleration of 5 m/s² for 5 seconds, use the formula:

$$Displacement=ut+1/2a{t}^2$$

Where,

• u = initial velocity (in this case, u = 0 m/s since the particle starts from rest)
• t = time (in this case, t = 5 s)
• a = acceleration (in this case, a = 5 m/s²)

By substituting the values, we get:

$$Displacement=0\ast 5+1/2\ast 5\ast 5^2$$ $$Displacement=0+125$$ $$Displacement=125m$$

Therefore, the particle moves 125 meters in 5 seconds.

CBSE Board Level:

1. To calculate the average speed of a car, use the formula:

$$AverageSpeed=Distance/Time$$

Where,

• Distance = 120 km
• Time = 2 hours (which must be converted to seconds for consistency: 2 * 60 * 60 = 7200 seconds)

By substituting the values, we get:

$$AverageSpeed=\frac{120000m}{7200s}$$ $$AverageSpeed=\frac{50}{3}m/s$$

Therefore, the average speed of the car is 50/3 m/s or approximately 66.67 km/h.

2. To find the average speed of a train, use the formula:

$$Speed=Distance/Time$$

Where,

• Distance = 300 km
• Time = 4 hours (which must be converted to seconds for consistency: 4 * 3600 = 14400 seconds)

By substituting the values, we get:

$$Speed=\frac{300000m}{14400s}$$ $$Speed=\frac{25}{3}m/s$$

Therefore, the average speed of the train is 25/3 m/s.

3. To calculate the maximum height reached by a ball thrown vertically upwards, use the formula:

$$Maximumheight=(u^2/2g)$$

Where,

• u = initial velocity (in this case, u = 15 m/s)
• g = acceleration due to gravity (in this case, g = 9.8 m/s²)

By substituting the values, we get:

$$Maximumheight=\frac{{15}^2}{2\times 9.8}$$ $$Maximumheight=\frac{225}{19.6}$$ $$Maximumheight\approx 11.48m$$

Therefore, the ball reaches a maximum height of approximately 11 meters.

4. To find the velocity with which a stone hits the ground, use the formula:

$$Finalvelocity=\surd (u^2+2gh)$$

Where,

• u = initial velocity (in this case, u = 0 m/s since the stone is dropped)
• g = acceleration due to gravity (in this case, g = 9.8 m/s²)
• h = height from which the stone is dropped (in this case, h = 50 m)

By substituting the values, we get:

$$Finalvelocity=\surd (0^2+2\times 9.8\times 50)$$ $$Finalvelocity=\surd (980)$$ $$Finalvelocity\approx 31.3m/s$$

Therefore, the stone hits the ground with a velocity of approximately 31.3 m/s.

5. To find the distance covered by a particle moving with a constant acceleration of 2 m/s² for 10 seconds, use the formula:

$$Displacement=ut+1/2a{t}^2$$

Where,

• u = initial velocity (in this case, u = 0 m/s since the particle starts from rest)
• t = time (in this case, t = 10 s)
• a = acceleration (in this case, a = 2 m/s²)

By substituting the values, we get:

$$Displacement=0\ast 10+1/2\ast 2\ast {10}^2$$ $$Displacement=0+100$$ $$Displacement=100m$$

Therefore, the particle covers 100 meters in 10 seconds.