Sure! Below is a list of tough questions related to motion along with their answers.
Tough Questions on Motion
-
Question 1: A car accelerates from rest at a constant rate of \(2 , \text{m/s}^2\). How far does it travel in the first 10 seconds?
-
Question 2: A projectile is launched vertically upward with an initial velocity of \(30 , \text{m/s}\). Calculate the maximum height reached.
-
Question 3: Two objects A and B are thrown vertically upwards with initial velocities of \(20 , \text{m/s}\) and \(10 , \text{m/s}\), respectively. Determine the time at which they collide.
-
Question 4: An object moves in a circular path with a radius of \(5 , \text{m}\) at a constant speed of \(10 , \text{m/s}\). What is the centripetal acceleration?
-
Question 5: A train moving with a speed of \(36 , \text{km/h}\) applies brakes and comes to a stop in \(5 , \text{s}\). What is the train’s deceleration?
-
Question 6: A body is thrown horizontally from a height of \(80 , \text{m}\). How far does it travel horizontally before hitting the ground? (Assume \(g = 9.8 , \text{m/s}^2\))
-
Question 7: A cyclist travels \(200 , \text{m}\) in the first \(10 , \text{s}\) and \(350 , \text{m}\) in the next \(15 , \text{s}\). Calculate the average speed for the entire journey.
Answers
-
Answer 1: \[ d = \frac{1}{2} a t^2 = \frac{1}{2} \times 2 , \text{m/s}^2 \times (10 , \text{s})^2 = 100 , \text{m} \]
-
Answer 2: \[ v^2 = u^2 + 2as \implies 0 = (30 , \text{m/s})^2 - 2 \cdot 9.8 , \text{m/s}^2 \cdot h \implies h = \frac{(30)^2}{2 \cdot 9.8} \approx 45.9 , \text{m} \]
-
Answer 3: \[ t_A = \frac{20}{9.81} \approx 2.04 , \text{s}, \quad t_B = \frac{10}{9.81} \approx 1.02 , \text{s} \] Since both are thrown at the same time, they will collide at the time when they are at the same height, approximately around \(1.02 , \text{s}\), before A reaches maximum height.
-
Answer 4: \[ a_c = \frac{v^2}{r} = \frac{(10 , \text{m/s})^2}{5 , \text{m}} = 20 , \text{m/s}^2 \]
-
Answer 5: \[ a = \frac{v_f - v_i}{t} = \frac{0 - 36 \times \frac{1000}{3600}}{5} \approx -2 , \text{m/s}^2 \]
-
Answer 6: \[ t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 80}{9.8}} \approx 4.04 , \text{s} \] If the horizontal speed is \(v\), then horizontal distance: \[ \text{Distance} = v \cdot t \]
-
Answer 7: \[ \text{Total distance} = 200 + 350 = 550 , \text{m}, \quad \text{Total time} = 10 + 15 = 25 , \text{s} \] \[ \text{Average speed} = \frac{550}{25} = 22 , \text{m/s} \]
Feel free to ask more about any specific question or topic!