Video Transcript
A particle starting from rest began moving in a straight line. Its acceleration 𝑎, measured in meters per second squared, and the distance 𝑥 from its start point, measured in meters, satisfy the equation 𝑎 is equal to 𝑥 squared over 15. Find the speed 𝑣 of the particle when 𝑥 equals 11 meters.
In this question, we are given the acceleration of a particle moving in a straight line as a function of displacement. 𝑎 is equal to 𝑥 squared over 15. Since the particle started from rest, we know that 𝑣 sub 𝑖 equals zero when 𝑥 sub 𝑖 equals zero, where 𝑣 sub 𝑖 is the initial velocity and 𝑥 is the distance measured from the start point. We know that integrating the acceleration 𝑎 of 𝑥 with respect to 𝑥 is equal to a half 𝑣 squared plus 𝑐, where 𝑐 is a constant of integration. When we integrate the expression between two limits, this is then equal to a half 𝑣 squared between the limits 𝑣 sub 𝑖 and 𝑣 sub 𝑓, where 𝑣 sub 𝑓 is the final velocity.
In this question, we need to calculate the speed of the particle 𝑣 when 𝑥 equals 11 meters. Since the acceleration is always positive, the velocity will not change direction. And the velocity in this question is therefore the same as the speed. Substituting in the values from this question, the left-hand side of our equation becomes the integral of 𝑥 squared over 15 with respect to 𝑥 between zero and 11. The right-hand side is equal to a half 𝑣 squared between zero and 𝑣 sub 𝑓, which we are trying to calculate. Integrating 𝑥 squared over 15 gives us 𝑥 cubed over 45. Substituting in our limits on the right-hand side, we have a half 𝑣 sub 𝑓 squared minus a half multiplied by zero squared. And the second part of this expression is zero.
Substituting in the limits on the left-hand side, we have 11 cubed over 45 is equal to a half 𝑣 sub 𝑓 squared. We can then multiply through by two such that 𝑣 sub 𝑓 squared is equal to 2662 over 45. Square rooting both sides and since 𝑣 sub 𝑓 must be positive, this is equal to 11 multiplied by the square root of 110 divided by 15. The speed of the particle 𝑣 when 𝑥 equals 11 meters is therefore equal to 11 multiplied by the square root of 110 over 15 meters per second, which is approximately equal to 7.69 meters per second.