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Semenov [28]
3 years ago
7

Present at least one example that illustrates acceleration. Is this a scalar or vector quantity? Explain why. ...?

Physics
2 answers:
Tatiana [17]3 years ago
8 0

Answer:

Vector

Explanation:

Acceleration is defined as the rate of change of velocity. As velocity is a vector quantity, so acceleration is also a vector quantity.

If a car starts from rest and attains some velocity after some time, then car is accelerating and the firection of acceleration is same as the direction of velocity.

If a car is moving and after applying the brakes it come to rest it means the motion of car has negative acceleration which means the direction of velocity and the direction of acceleration is opposite to each other.

SIZIF [17.4K]3 years ago
5 0
Acceleration is a vector because it has magnitude and direction

Example :
a Car stopping at the rate of 5 m/s. In this example, the direction of acceleration is the opposite of the direction of velocity, and the magnitude is 5 m/s

Hope this helps
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2 years ago
A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre
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Answer:

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Explanation:

We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:

y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}

v_1=v_{01}+a_1t

We must consider that it's launched from the ground (y_{01}=0m) and from rest (v_{01}=0m/s), with an upwards acceleration a_{1}=28m/s^2 that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m

And the velocity achieved in part 1:

v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 (y_{02}=1317.26m) and its initial velocity is the one achieved in part 1 (v_{02}=271.6m/s), now in free fall, which means with a downwards acceleration a_{2}=-9,8m/s^2. For the data we have it's faster to use the formula v_f^2=v_0^2+2ad, where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

v_{02}^2+2a_2(y_2-y_{02})=v_2^2=0m/s

Then, to get y_2, we do:

2a_2(y_2-y_{02})=-v_{02}^2

y_2-y_{02}=-\frac{v_{02}^2}{2a_2}

y_2=y_{02}-\frac{v_{02}^2}{2a_2}

And we substitute the values:

y_2=y_{02}-\frac{v_{02}^2}{2a_2}=(1317.26m)-\frac{(271.6m/s)^2}{2(-9.8m/s^2)}=5080.86m

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Explanation: I HOPE THAT HELPED

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