Question
Distance traveled by a particle
Original question: Q3 Distance traveled by a particle
A particle travels in a straight line with a constant acceleration of 3 meters per second per second. If the velocity of the particle is 10 meters per second at time 2 seconds, how far does the particle travel during the time interval when its velocity increases from 4 meters per second to 10 meters per second?
A 20 m B 14 m C 7 m D 6 m
Related Con... Topic 8.2 Skill 1.D
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Expert Verified Solution
Key takeaway: With constant acceleration, the cleanest approach is to use the kinematics relation between speed and displacement.
Step 1: Use the constant-acceleration formula
For motion with constant acceleration,
Here:
- initial velocity
- final velocity
- acceleration
Step 2: Solve for displacement
=\frac{10^2-4^2}{2(3)} =\frac{100-16}{6} =\frac{84}{6} =14.$$ So the particle travels **$\boxed{14\text{ m}}$** during that interval. **Correct option: B** --- **Pitfalls the pros know** 👇 A frequent mistake is to use average velocity without checking that the acceleration is constant. The formula $v^2=u^2+2as$ is the fastest and most reliable method here. **What if the problem changes?** If the question asked for the time taken instead of the distance, you could use $v=u+at$: $$t=\frac{10-4}{3}=2\text{ s}.$$ Then the distance could also be found from $s=ut+\tfrac12at^2$ and would still give 14 m. `Tags`: constant acceleration, kinematics, displacementFAQ
How far does the particle travel while its velocity increases from 4 m/s to 10 m/s?
Using $v^2=u^2+2as$ with $u=4$, $v=10$, and $a=3$, we get $s= rac{100-16}{6}=14$. The particle travels 14 m.
Which kinematics formula is used here?
The constant-acceleration equation $v^2=u^2+2as$ is used because the acceleration is given as constant and the problem asks for distance traveled between two speeds.