Question

A bicycle is moving in a straight line PQ with a constant acceleration. The velocity of the bicycle at point P and Q is 15ms-1 and 25ms-1 respectively. What is the acceleration of the bicycle if the distance between P and Q is 40m?

Answer 1

To solve this problem, we can use the kinematic equation that relates velocity, acceleration, and distance:

$v^2=u^2+2as$

where: v = final velocity u = initial velocity a = acceleration s = distance

Given that the velocity at point P (initial velocity) is 15 m/s and the velocity at point Q (final velocity) is 25 m/s, and the distance between P and Q is 40 m, we can substitute these values into the equation.

At point P: $u=15\text{m/s}$

At point Q: $v=25\text{m/s}$

The distance between P and Q: $s=40\text{m}$

Substituting these values into the equation, we get: $25^2=15^2+2a(40)$

Simplifying the equation: $625=225+80a$

Rearranging the equation to solve for acceleration: $80a=625-225$ $80a=400$

Dividing both sides of the equation by 80, we find: $a=\frac{400}{80}$ <m