Shortcut Methods
Shortcut Methods and Tricks to Solve Numerical Problems on Motion in a Straight Line
JEE Main Level:
- To find the distance traveled by a car in 10 seconds, use the formula:
$$s = ut + 1/2 at^2$$
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:
$$s = 0 * 10 + 1/2 * 2 * 10^2$$ $$s = 100 m$$ Therefore, the car travels a distance of 100 meters in 10 seconds.
- 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:
$$Speed = \frac{200000 m}{7200s}$$ $$Speed = \frac{50}{3} m/s$$
Therefore, the average speed of the train is 50/3 m/s.
- To find the maximum height reached by a ball thrown vertically upwards, use the formula:
$$Maximum height = (u^2 / 2g)$$
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:
$$Maximum height = (20^2 / 2 \times 9.8)$$ $$Maximum height = \frac{400}{19.6}$$ $$Maximum height = 20.45 m$$
Therefore, the ball reaches a maximum height of approximately 20 meters.
- To find the velocity with which a stone hits the ground, use the formula:
$$Final velocity = √(u^2 + 2gh)$$
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:
$$Final velocity = √(0^2 + 2 \times 9.8 \times 100)$$ $$Final velocity = √(1960)$$ $$Final velocity = 44.3 m/s$$
Therefore, the stone hits the ground with a velocity of approximately 44.3 m/s.
- To find the displacement of a particle moving with a constant acceleration of 5 m/s² for 5 seconds, use the formula:
$$Displacement = ut + 1/2 at^2$$
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:
$$Displacement = 0 * 5 + 1/2 * 5 * 5^2$$ $$Displacement = 0 + 125$$ $$Displacement = 125 m$$
Therefore, the particle moves 125 meters in 5 seconds.
CBSE Board Level:
- To calculate the average speed of a car, use the formula:
$$Average Speed = Distance/Time$$
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:
$$Average Speed = \frac{120000 m}{7200s}$$ $$Average Speed = \frac{50}{3} m/s$$
Therefore, the average speed of the car is 50/3 m/s or approximately 66.67 km/h.
- 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:
$$Speed = \frac{300000 m}{14400s}$$ $$Speed = \frac{25}{3} m/s$$
Therefore, the average speed of the train is 25/3 m/s.
- To calculate the maximum height reached by a ball thrown vertically upwards, use the formula:
$$Maximum height = (u^2 / 2g)$$
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:
$$Maximum height = \frac{15^2}{2 \times 9.8}$$ $$Maximum height = \frac{225}{19.6}$$ $$Maximum height \approx 11.48 m$$
Therefore, the ball reaches a maximum height of approximately 11 meters.
- To find the velocity with which a stone hits the ground, use the formula:
$$Final velocity = √(u^2 + 2gh)$$
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:
$$Final velocity = √(0^2 + 2 \times 9.8 \times 50)$$ $$Final velocity = √(980)$$ $$Final velocity \approx 31.3 m/s$$
Therefore, the stone hits the ground with a velocity of approximately 31.3 m/s.
- To find the distance covered by a particle moving with a constant acceleration of 2 m/s² for 10 seconds, use the formula:
$$Displacement = ut + 1/2 at^2$$
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:
$$Displacement = 0 * 10 + 1/2 * 2 * 10^2$$ $$Displacement = 0 + 100$$ $$Displacement = 100 m$$
Therefore, the particle covers 100 meters in 10 seconds.
