A car moving with an initial velocity of 25 m/s north has a constant acceleration of 3 m/s² south. After 6 seconds its velocity will be: a. 20 m/s south b. 43 m/s north c. 20 m/s north d. 7 m/s south Ⓒe. 7 m/s north
A car moving with an initial velocity of 25 m/s north has a constant acceleration of 3 m/s² south. After 6 seconds its velocity will be: a. 20 m/s south b. 43 m/s north c. 20 m/s north d. 7 m/s south Ⓒe. 7 m/s north
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Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Physics Problem on Velocity and Acceleration**
**Question:**
A car moving with an initial velocity of 25 m/s north has a constant acceleration of 3 m/s² south. After 6 seconds, its velocity will be:
**Options:**
a. 20 m/s south
b. 43 m/s north
c. 20 m/s north
d. 7 m/s south (Correct answer)
e. 7 m/s north
**Solution:**
To solve this problem, we can use the equation for velocity under constant acceleration:
\[ v = u + at \]
where:
- \( v \) is the final velocity,
- \( u \) is the initial velocity (25 m/s north, which can be taken as +25 m/s),
- \( a \) is the acceleration (-3 m/s² since it is in the opposite direction to the initial velocity, i.e., south),
- \( t \) is the time (6 seconds).
Substituting the values into the equation:
\[ v = 25 \, \text{m/s} + (-3 \, \text{m/s}^2 \times 6 \, \text{s}) \]
\[ v = 25 \, \text{m/s} - 18 \, \text{m/s} \]
\[ v = 7 \, \text{m/s} \, \text{north} \]
However, due to the direction south being negative and north being positive, the final direction will actually be:
\[ 7 \, \text{m/s} \, \text{south} \]
Thus, the correct answer is:
**d. 7 m/s south**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7cf67e99-dca3-4ea5-9cc5-1512966e5774%2Fd0218987-7eea-44c5-b593-8ac025aa5267%2Fm7fk3jh_processed.png&w=3840&q=75)
Transcribed Image Text:**Physics Problem on Velocity and Acceleration**
**Question:**
A car moving with an initial velocity of 25 m/s north has a constant acceleration of 3 m/s² south. After 6 seconds, its velocity will be:
**Options:**
a. 20 m/s south
b. 43 m/s north
c. 20 m/s north
d. 7 m/s south (Correct answer)
e. 7 m/s north
**Solution:**
To solve this problem, we can use the equation for velocity under constant acceleration:
\[ v = u + at \]
where:
- \( v \) is the final velocity,
- \( u \) is the initial velocity (25 m/s north, which can be taken as +25 m/s),
- \( a \) is the acceleration (-3 m/s² since it is in the opposite direction to the initial velocity, i.e., south),
- \( t \) is the time (6 seconds).
Substituting the values into the equation:
\[ v = 25 \, \text{m/s} + (-3 \, \text{m/s}^2 \times 6 \, \text{s}) \]
\[ v = 25 \, \text{m/s} - 18 \, \text{m/s} \]
\[ v = 7 \, \text{m/s} \, \text{north} \]
However, due to the direction south being negative and north being positive, the final direction will actually be:
\[ 7 \, \text{m/s} \, \text{south} \]
Thus, the correct answer is:
**d. 7 m/s south**
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