Question
At a certain time a particle had a speed of $18 \mathrm{~m} / \mathrm{s}$ in the positive $x$ direction, and $2.4 \mathrm{~s}$ later its speed was $30 \mathrm{~m} / \mathrm{s}$ in the opposite direction. What is the average acceleration of the particle during this $2.4 \mathrm{~s}$ interval?
Step 1
The initial velocity \( v_i \) is \( 18 \, \text{m/s} \) in the positive \( x \) direction. The final velocity \( v_f \) is \( 30 \, \text{m/s} \) in the opposite direction, which means it is \( -30 \, \text{m/s} \). Show more…
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Acceleration At a certain time a particle had a speed of 18 $\mathrm{m} / \mathrm{s}$ in the positive $x$ direction, and 2.4 s later its speed was 30 $\mathrm{m} / \mathrm{s}$ in the opposite direction. What is the average acceleration of the particle during this 2.4 s interval?
A particle had a velocity of $18 \mathrm{~m} / \mathrm{s}$ in the $+x$ direction and 2.4 s later its velocity was $30 \mathrm{~m} / \mathrm{s}$ in the opposite direction. What was the average acceleration of the particle during this 2.4-s interval?
A particle had a speed of $18 \mathrm{~m} / \mathrm{s}$ at a certain time, and $2.4 \mathrm{~s}$ later its speed was $30 \mathrm{~m} / \mathrm{s}$ in the opposite direction. What were the magnitude and direction of the average acceleration of the particle during this $2.4$ s interval?
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