Uniformly Accelerated Motion Bc829C

1. **Problem 1:** A car starts from rest and accelerates uniformly at $3\ \text{m/s}^2$ for $5\ \text{s}$. Find the final velocity and the distance covered. 2. **Formula used:** - Final velocity: $$v = u + at$$ - Distance covered: $$s = ut + \frac{1}{2}at^2$$ 3. **Given:** - Initial velocity, $u = 0\ \text{m/s}$ (starts from rest) - Acceleration, $a = 3\ \text{m/s}^2$ - Time, $t = 5\ \text{s}$ 4. **Calculate final velocity:** $$v = 0 + 3 \times 5 = 15\ \text{m/s}$$ 5. **Calculate distance covered:** $$s = 0 \times 5 + \frac{1}{2} \times 3 \times 5^2 = \frac{1}{2} \times 3 \times 25 = 37.5\ \text{m}$$ --- 6. **Problem 2:** A ball is thrown vertically upward with an initial velocity of $20\ \text{m/s}$. Calculate the time taken to reach the highest point and the maximum height. 7. **Formula used:** - Time to highest point: $$t = \frac{v - u}{a}$$ (Here, $v=0$ at highest point, $a = -9.8\ \text{m/s}^2$ due to gravity) - Maximum height: $$h = ut + \frac{1}{2}at^2$$ 8. **Given:** - Initial velocity, $u = 20\ \text{m/s}$ - Final velocity at highest point, $v = 0\ \text{m/s}$ - Acceleration, $a = -9.8\ \text{m/s}^2$ 9. **Calculate time to highest point:** $$t = \frac{0 - 20}{-9.8} = \frac{-20}{-9.8} \approx 2.04\ \text{s}$$ 10. **Calculate maximum height:** $$h = 20 \times 2.04 + \frac{1}{2} \times (-9.8) \times (2.04)^2$$ $$h = 40.8 - \frac{1}{2} \times 9.8 \times 4.16 = 40.8 - 20.4 = 20.4\ \text{m}$$ --- 11. **Problem 3:** A train moving at $30\ \text{m/s}$ applies brakes and comes to rest in $10\ \text{s}$. Find the acceleration and distance traveled during braking. 12. **Formula used:** - Acceleration: $$a = \frac{v - u}{t}$$ - Distance: $$s = ut + \frac{1}{2}at^2$$ 13. **Given:** - Initial velocity, $u = 30\ \text{m/s}$ - Final velocity, $v = 0\ \text{m/s}$ - Time, $t = 10\ \text{s}$ 14. **Calculate acceleration:** $$a = \frac{0 - 30}{10} = -3\ \text{m/s}^2$$ 15. **Calculate distance:** $$s = 30 \times 10 + \frac{1}{2} \times (-3) \times 10^2$$ $$s = 300 - \frac{1}{2} \times 3 \times 100 = 300 - 150 = 150\ \text{m}$$