Question

A cyclist traveling at 30.0m/s along a straight road comes uniformly to stop in 5.00s. Determine the stopping acceleration, the stopping distance.

Answer 1

Answer:

I. Stopping acceleration = 6 m/s²

II. Stopping distance, S = 75 meters

Explanation:

Given the following data;

Final velocity = 30 m/s

Time = 5 seconds

To find the stopping acceleration;

Mathematically, acceleration is given by the equation;

$Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}$

Substituting into the equation;

$a = \frac{30 - 0}{5}$

$a = \frac{30}{5}$

Acceleration = 6 m/s²

II. To find the stopping distance, we would use the third equation of motion;

$V^{2} = U^{2} + 2aS$

Where;

V represents the final velocity measured in meter per seconds.

U represents the initial velocity measured in meter per seconds.

a represents acceleration measured in meters per seconds square.

Substituting into the equation, we have;

30² = 0² + 2*6*S

900 = 12S

S = 900/12

S = 75 meters