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  1. ...time . 3. First statment is false. The acceleration at any time and position is equal to the acceleration due to gravity, which is not zero.Second...

  2. You have some information missing. How much does its velocity decrease? As far as I remember this question, the velocity should decrease from 30 m/s to 20 m/s in 100 m. So, I will use this to do the question. Initial velocity, u = 30 m/s Final...

  3. Time needed for reaching the MAXIMUM velocity : u=0, v=10ms^-1, a=3.3ms^-2, By v=u+at,10=0+3.3t t=10/3.3 s distance of the runner's velocity change from 0 to 10ms^-1 : By v^2-u^2=2aS, 10^2-0=2(3.3)S S=100/6...

  4. (a) Distance covered by satellite in one revolution round the earth = 2 x pi x (350+6370) km where pi = 3.14159..... Hence, average speed = distance/time = 2 x pi x (350+6370) /1.5 km/h (b) Similar, distance covered by satellite in one revolution = 2 x pi x (h+ 6370) km where h (in km) is the...

  5. Let d be the distance between A and B hence time t1 to travel from A to B = distance/speed = d/1 = d time t2 to travel from B to A = d/2 Total distance travelled in the whole journey = 2d total time taken = t1 + t2 = d + d/2 = 3d/2 Therefore, average...

  6. Answer : E 任何物件於地球表面,都受到gravitational force (g) ge 影響。Gravitation force會向住地心 (earth core) 拉住件物件。 而你亦都應該學左 Gravitation force = mg 所有物件都有一個Gravitation force (mg) 向下拉 如果該物件係置於地面上...

  7. s=ut+(1/2)at2 if t=1, s=0+(1/2)(10)(1)2 s=5 m if t=2, s=0+(1/2)(10)(2)2 s=20 m if t=3, s=0+(1/2)(10)(3)2 s=45 m 45 m係指該物件由t=0到t=3期間所行走的distance/displacement 0 s to 1 s is the first second 1.01 s to 2 s is the second second(第二秒) 2.01 s to 3 s is...

  8. Initial velocity, u = 50 km/h = 50 / 3.6 = 13.89 m/s Final velocity, v = 0 Stopping distance, s = 15 m By equation of motion, v2 = u2 - 2as 0 = (13.89)2 - 2a(15) Deceleration, a = 6.43 ms-2 Now, initial velocity, u = 70 km/h = 70/3.6 = 19.44 m/s By equation of motion, v2 = u2...

  9. First of all, after 20 minutes, the car has travelled 60 x (20/60) = 20 km. So at that time the car still has to travel 40km. If the car travels at 80 km/h for the remaining distance, it needs a further 40/80 = 0.5 hours or 30...

  10. ...some celestial body as "stationary" and describe others' position relative to the reference point. To learn more and in depth, you may borrow To...