EK physics 1001 # 57

[Temperature101]

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A particle moving at 5 m/s reverses its direction in 1 second to move at 5 m/s in the opposite direction. If its acceleration is constant, what is its speed at 0.5 s ?

A. 0 m/s
B. 1.25 m/s
C. 2.5 m/s
D. 5 m/s

The answer is A ...Can someone explain?

 

[stsa84]

Let's say the particle is initially moving to the left.

It is moving to the left at 5 m/s, and 1 second later, it is moving at 5 m/s to the right. Since the acceleration is constant, you can easily see that at 0.5 seconds, the particle is moving at 0 m/s. It takes 0.5 seconds to come to a stop, and 0.5 seconds to return to 5 m/s going the opposite direction.

From a formula point of view:

v(final) = v(initial) + at

v(initial) = 5 m/s
a = -10 m/s^2 (a = change in velocity/time = (-5 - (5)) / 1 = -10)
t = 0.5 s

v(final) = 5m/s + (-10 m/s^2)(0.5) = 0 m/s

 

[Temperature101]

Let's say the particle is initially moving to the left.

It is moving to the left at 5 m/s, and 1 second later, it is moving at 5 m/s to the right. Since the acceleration is constant, you can easily see that at 0.5 seconds, the particle is moving at 0 m/s. It takes 0.5 seconds to come to a stop, and 0.5 seconds to return to 5 m/s going the opposite direction.

From a formula point of view:

v(final) = v(initial) + at

v(initial) = 5 m/s
a = -10 m/s^2 (a = change in velocity/time = (-5 - (5)) / 1 = -10)
t = 0.5 s

v(final) = 5m/s + (-10 m/s^2)(0.5) = 0 m/s

Thanks...I get it from the formula point of view. Can you explain more why you said it takes 0.5s for the particle to come to a stop due to constant acceleration?

 

[stsa84]

New scenario. Say you are walking forwards at a rate of 2 steps per second. You then undergo an acceleration of -1 steps per second. How many seconds would it take for your velocity to reach 0 steps per second? How many seconds to reach 2 steps per second, going backwards?

Since the acceleration is constant, your velocity will change by the exact same quantity, every single second. In other words, your velocity will decrease by 1 step per second, every second. At some point you will stop, and at another point you will begin moving backwards.

At time = 0s, you have a velocity of 2 steps/second, forwards
At time = 1s, you have a velocity of 1 step/second, forwards
At time = 2 s, you have a velocity of 0 steps/second
At time = 3s, you have a velocity of 1 step/second, backwards
At time = 4s, you have a velocity of 2 steps/second, backwards


In the problem you first mentioned, it takes 1 full second to go from +5 m/s to -5m/s.

t = 0s, +5m/s
t = 1s, -5m/s

t = 0.5s, 0 m/s

Every second, the particle's acceleration decreases by 10m/s. After 0 seconds, the velocity is unchanged, after one second, it has decreased by 10 m/s to -5m/s. At half of a second, it has decreased by only half of the amount it would decrease in a second, namely, by 5m/s, so the current velocity at 0.5s is 0m/s.

 

[MT_15]

Conceptually, a better example would have been an object tossed upward reaching a velocity of zero because of the constant downward acceleration due to gravity.