A car moving with an initial velocity of 25 mis north has a constant acceleration of 3 m/s2 south. After 6 s its velocity will be:
Correct answer: A. 7 m/s north
- A. 7 m/s north
- B. 43 m/s north
- C. 7 m/s south
- D. 20 m/s north
Explanation
The final velocity of the car can be calculated using the equation: vf = vi + a·t, where vi is the initial velocity, a is the acceleration, and t is the time.Given: vi = 25 m/s north, a = -3 m/s2 (since it's south), and t = 6 s.Substitute these values into the formula: vf = 25 + (-3) * 6 = 25 - 18 = 7 m/s north.Therefore, the correct answer is 7 m/s north.Option B assumes an increase in velocity, which is incorrect because the acceleration is in the opposite direction. Option C suggests a reversal in direction, which hasn't occurred in the given time frame. Option D underestimates the effect of the opposing acceleration.
Report an error
The more specific you are, the faster it gets fixed. A source beats an opinion.
Prefer email? [email protected]
About Force and Motion
Displacement, velocity and acceleration describe motion, including the horizontal and vertical components of projectile motion. The chapter connects these quantities with Newton's laws, momentum and impulse, then applies conservation of momentum to collisions, distinguishing elastic collisions from inelastic ones.
Practise Force and Motion
1,416 free Force and Motion MCQs from Physics, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Discussion
Stuck on an option, or know a faster way to get there? Ask or explain it here.
No comments yet. Be the first to explain this one.
Exams that ask Physics questions like this
Physics is on 16 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.
More Force and Motion questions
A cricket ball is hit so that it travels straight up in air and it acquires 3 s to reach the maximum height. Its initial velocity is:
A body is thrown with the velocity (4i+3j)m/s. Its maxium height is (g=10m/s2)
A block of mass 1 kg moves along a frictionless surface with a velocity of 10 ms-1 and it colides with a stationary block of mass 10 kg. After the collision, the first block rebounds with a velocity of 5 ms-1 and the other one moves off with a velocity of 1.5ms-1.
Time displacement (t-x) graph of two objects A and B is shown in Fig. The ration of their speeds [VA/VB] is (tan 37o = 3/4)
A stone is dropped from a cliff. The graph (carefully note the axes) which has represents its motion while it falls is:
A 1.0 kg ball moving at 2.0m/s perpendicular to a wall rebounds from the wall at 1.5m/s. The change in the momentum of the ball is: