This quiz contains practice questions for Speed, Velocity & Acceleration (O Level). There are explanations for some of the questions after you submit the quiz. If you could not figure out why a particular option is the answer, feel free to drop a comment below or ask a question in ‘O’ & ‘A’ Level Discussion section of the forum. If you wish to contribute questions to this quiz, please contact us.

The quiz comprises of 10 questions, which are randomly selected from a large pool of questions. After completion, you can refresh the page and try the quiz again for a new set of 10 questions.

Back To List of O Level Physics Quiz

7. A ball is thrown vertically upwards at an initial speed of 23 \, \text{m s}^{-1} . What is the distance traveled by the ball 3 seconds after it was thrown?

29 mcorrect

55 m

24 mwrong

50 m

Can someone please explain why the answer is 29 and not 24?

THE ANSWER IS 24

Note that the acceleration due to gravity is $10 \, \text{m s}^{2}$. The ball would have reached zero velocity in

lessthan 3 seconds.You will need to break the problem into parts and not solve it with the application of one single kinematics equation.

lun khao

answer this please with calculation:A car travels at a constant speed of 5 \, \text{m s}^{-1} for 2 seconds, then accelerates at a rate of 4 \, \text{m s}^{-2} for 3 seconds. What is the total distance traveled by the car?

70 m

“A ball is thrown vertically upwards off a cliff at a speed of 23ms^-1. It falls to the bottom of the cliff after 10 seconds. What is the height of the cliff? ( g = 10ms^-2 )”

The Answer: 130 m

If I were to use s = ut + 1/2(at^2) , I would get the answer: -270m , which is not correct according to the answer. (this can be negative as it is a displacement value)

s: height (?), u: initial velocity (23), t: time (10), a: acceleration (-10 due to gravity)

If I were to use s = 1/2(u + v)t , I would get the answer 115m, which is not correct according to the answer.

s: height (?), u: initial velocity (23), v: final velocity (0)(because the ball has fallen to the bottom and is presumably on the floor), t: time (10)

If I were to use v^2 = u^2 + 2as, and rearrange to get s = (v^2 – u^2)/2a , I would get the answer: -26.45m , which is not correct according to the answer. (this can be negative as it is a displacement value)

s: height (?), u: initial velocity (23), v: final velocity (0)(because the ball has fallen to the bottom and is presumably on the floor), a: acceleration (-10 due to gravity)

(I am using the velocity as 23 throughout this even though it says that 23 is the ‘speed’, i’m not sure if that is where I am going wrong because then I would have to find the velocity by using v = s/t , but I don’t know s, and I am sure that here; ‘speed’ is interchangeable with ‘velocity’)

I don’t know where I am going wrong here, whether or not I am messing up somewhere with the wrong equation, or the final velocity. I am just so stuck on this for some reason, I would appreciate any help ASAP.

Hello. Your website is very helpful. Can you please explain why the answer to the following question is the last option?

8. A ball is thrown vertically upwards. It lands on the top of a cliff after attaining a maximum height of 100 m in 7 s. What is the height of the cliff if the ball reaches the top of the cliff after 9 s? What is the height of the cliff? $\left( g = 10 \, \text{m s}^{ – 2} \right)$

ball thrown onto cliff

70 m

90 m

60 m

80 m

Thank you.

Use the three kinematics equations (located here).

Step 1: Break the problem into three parts:

– Ball at t = 0

– Ball at t = 7s

– Ball at t = 9s

Step 2: Focus on the t = 7 s and t = 9s.

Step 3: Use the kinematics equation to find the distance travelled by the ball from t = 7s to t = 9s.

(Hint: The initial velocity will be zero, time will be 2 seconds, acceleration will be the gravitational acceleration and s will be the distance travelled)

The hight of cliff after 9s will be 80m.

Pls help me. An object accelerates from 3m/s to 10 m/s in 3s. Find distance travelled.

Please refer to this post: Acceleration (O Level). In particular, pay close attention to the kinematics equations.

If you require more help, you can post on the forum

the answer to this question is 2.33 m if I am not wrong?

Nope, that is not correct.

There are two methods to solving this question:

1) Use one of the kinematics equations; or

2) Plot out the velocity-time graph of the object. (Refer to Reading Kinematics Graphs

Yeah. I’ve shown all the correct calculations on “Khamir Kheir’s” reply section. Please check and correct if any mistakes. Thank you!

Given

Intial velocity = 3m/s

Final velocity = 10m/s

Time = 3s

a = v – u/t

= 10 -3/3

= 7/3 = 2.33

correct

This is wrong. The question asked to find the distance but you found acceleration instead.

We can use the basic one of the four equations of motion.

Given,

initial velocity = 3m/s

final velocity = 10m/s

time = 3s

distance = ?

s = (u+v/2) t

(where, s = distance, u = initial velocity, v = final velocity, t = time)

So,

s = (3+10/2) x 3

s = 6.5 x 3

s = 19.5m

The answer is correct but you will have to take note of the presentation of the equation. It should be:

$$s = \frac{1}{2}(u+v)t$$

s=1/2(u+v)t

= 0.5(3+10)3

= 19.5 metres

The distance travelled will be 21m.