The velocity, V feet per second, of a model rocket t seconds after

Question

﻿The velocity, V feet per second, of a model rocket t seconds after launch is given by V = -32t + C, where C is a positive constant. What is the velocity of the rocket 2 seconds after it was launched?

(1) The rocket reaches its maximum height and begins descending 1.5 seconds after it was launched.
(2) The rocket's initial velocity was 48 feet per second.

gmatophobia · Accepted Answer

Formula Given: V = -32t + C ; C is a positive number

Question: V at time = 2 seconds?

Inference : If we can find the value of C, we will be able to answer the question.

Statement 1

(1) The rocket reaches its maximum height and begins descending 1.5 seconds after it was launched.

At its maximum height, the velocity of the rocket is zero. Hence, we know that at t = 1.5, V = 0

0 = -32(1.5) + C

We can find the value of C. The statement alone is sufficient to answer the question. We can eliminate B, C, and E.

Statement 2

(2) The rocket's initial velocity was 48 feet per second.

At t = 0, V = 48

48 = -32(0) + C

C = 48

As we have the value of C, the statement alone is sufficient to answer the question.

Option D

JeffTargetTestPrep · Answer

We need to answer the question:

V = -32(2) + C = ? => C = ?

Statement One Alone:

=> The rocket reaches its maximum height and begins descending 1.5 seconds after it was launched.

According to the given velocity formula, the velocity decreases as time passes after launch. As long as its speed is positive, the rocket can continue to ascend. Once its velocity becomes zero, the rocket will reach its maximum height. So, we have:

-32(1.5) + C = 0

C = 48

Since we have a unique answer to the rephrased question, statement one is sufficient. Eliminate answer choices B, C, and E.

Statement Two Alone:

=> The rocket’s initial velocity was 48 feet per second.

Since, at the initial moment of the launch, t = 0, we have:

-32(0) + C = 48

C = 48

Since we have a unique answer to the rephrased question, statement two is sufficient.

Answer: D

HSWorNothing · Answer

I really like this question. However, I would also like to highlight that this may not be very representative of a GMAT question since to understand the significance of statement one, we would need to have a good understanding of the relationship between height/velocity (which arguably is more physics than math).

Let me know if my assessment is accurate/fair.

Let me know if my assessment is accurate/fair.

Bunuel · Answer

This is a GMAT Prep (Focus) question, so it is as representative of an actual GMAT as it gets. Questions from GMAT Prep were once used in the actual test.

GMATinsight · Answer

Answer: Option D

Please check the Video explanation here

https://www.youtube.com/watch?v=f730MFXZsJ8

Confide · Answer

Can anyone explain why the velocity is 0 at maximum height? An explanation would be appreciated rather than just stating it without any reasoning behind it. Thank you

HarshavardhanR · Answer

Perhaps this example will help.

Remember what happens when we throw a ball up in the air.

- Once we throw the ball up, it travels upward with some velocity.
- It reaches a certain maximum height.
- Then, it starts falling i.e., it travels downward with some velocity.

The key point is this:  At that exact maximum height, the ball actually  pauses  for the smallest of split-seconds, and only then moves downwards. In other words, at that  exact  point of max height,  the velocity of the ball is 0 .

___
Harsha

SergejK · Answer

but it is nowhere stated that the rocket is flying straight up. the rocket could actually be flying in a curve and reaching a maximum height would not necessarily mean that V=0. A bit imprecise, I think.

HarshavardhanR · Answer

To be honest, I agree with this take.

- While it  may  be inferable that we are dealing with vertical movement only, thanks to the given equation (32 ft/s2 is ---> 9.8 m/s2 ---> which is acceleration due to gravity, which only applies vertically downward ---> so, we must be dealing with purely vertical movement basis the equation), to be able to infer this requires some specific science knowledge beyond general knowledge. So,  the question should ideally specify vertical movement .

- It may be that the test-maker intent was to consider the general lay-person understanding ("rocket typically goes straight up"). In which case, the velocity is 0 at the point of maximum height.

- of course, if we consider the rocket to move along a curve (projectile motion), at the point of maximum height - 
---> Vertical velocity is 0.
---> Horizontal velocity exists (not 0).

So, strictly speaking,  overall  velocity is not 0. Agreed.

That said, I think the test-maker intent here is upward movement rather than movement along a curve.

___
Harsha

agrasan · Answer

Hi experts  KarishmaB   chetan2u   MartyMurray   GMATCoachBen

I am still not clear, why this (D) would be the answer here? I think (B) should be the answer.

Using my  understanding of Physics  here:
1. Velocity at the highest point can't necessarily be zero if rocket has both vertical and horizontal velocity. In that case Statement-1 would be Insufficient. 
2. "Rocket launched" can also happen in parabolic trajectory apart from vertical pose only. In former case, it will have both horizontal and vertical velocity components. In that case Statement-1 would be Insufficient.

Can you please let me know which part of the question guides us that we have to assume that rocket  has vertical velocity only ? I think due to this assumption, answer shown is (D).

KarishmaB · Answer

It is a model rocket so we can pretty much expect it to be basic - start with an initial velocity, go straight up and velocity reduces because of gravity. Finally reaches the top point when velocity becomes 0 and comes back down. The reason 'velocity' is mentioned - because it is negative while coming down as per the equation. No other reason.

ManishaPrepMinds · Answer

The clue is in the equation  V = −32t + C .
The constant −32 represents acceleration due to gravity in ft/s2, which applies only to  vertical motion .
At the highest point, vertical velocity = 0. The negative sign means downward motion by the usual sign convention (up = positive, down = negative).
If horizontal velocity were involved, we’d need separate equations for x- and y-components. Since only one equation is given, the question is about vertical velocity only.
Therefore, Statement (1) is sufficient, and the answer is  (D) .

MartyMurray · Answer

The question defines the velocity for us by saying the following:

The velocity, V feet per second, of a model rocket t seconds after launch is given by V = -32t + C, where C is a positive constant.

We see that, as defined by the question, the velocity of the rocket has only a vertical component. So, we need to be concerned with the vertical velocity only.

agrasan · Answer

Thanks  MartyMurray 
Did you identify that questions asks for vertical velocity only due to the co-efficient  -32  which can be due to gravity and other frictional forces?

MartyMurray · Answer

Yes, the presence of the single coefficient -32 indicates that only gravity is playing a role. If friction or horizontal velocity were taken into consideration, a more complex formula would be required for representing the trajectory of the rocket.

That said, I don't find this question to be a fair question because answering it correctly basically requires prior understanding of everything we're talking about, which is not what the GMAT is meant to test.

nandini14 · Answer

I do not understand why would we take v=0?

Bunuel · Answer

When the rocket goes up, it keeps slowing down until it reaches the very top, its maximum height. At that exact moment, it is not moving upward anymore, and it has not yet started moving downward. For that instant, its velocity is zero. That’s why in statement (1) we put v = 0 at t = 1.5 seconds.

Jbin96 · Answer

What I am having trouble understanding is sure we can answer for "C", but how does it relate to the original question or solving for 't' of 2 seconds? Just the fact we have C so we then plug in "2" for T?

Bunuel · Answer

Yes. That’s it. The whole point is that the formula already gives V in terms of t and C. Once you know C, you simply plug t = 2 into V = -32t + C and you get the velocity at 2 seconds. The statements are sufficient only because they let you find C.

The velocity, V feet per second, of a model rocket t seconds after

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The velocity, V feet per second, of a model rocket t seconds after

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