Question

In 1994, Tony Lang of the United States rode his motorcycle a short distance of 4.00 x 102 m in the short interval of 11.5 s. He started from rest and crossed the finish line with a speed of about 2.50 x 102 km/h. Find the magnitude of Lang’s acceleration as he traveled the 4.00 x 102 m distance.

Answer 1

The magnitude of Tony Lang's acceleration as he traveled the 4.00 x 10² m distance is approximately 6.038 m/s² after converting his final speed to meters per second and using the kinematic equation for uniformly accelerated motion.

Step-by-step explanation:

To find the magnitude of Tony Lang's acceleration, we first need to convert his final speed into meters per second. The final speed is given as 2.50 x 102 km/h, which we convert to 2.50 x 105 m/h. Then, we convert hours into seconds: 2.50 x 105 m/h * (1/3600) h/s = 69.44 m/s.

Now, we use the kinematic equation for uniformly accelerated motion without an initial velocity (since he started from rest):

• Final speed (v) = Initial speed (u) + acceleration (a) * time (t)
• 69.44 m/s = 0 + a * 11.5 s
• a = 69.44 m/s / 11.5 s
• a = 6.038 m/s2

Hence, the magnitude of Lang's acceleration over the 4.00 x 102 m distance was approximately 6.038 m/s2.