Question

A motorcycle traveling along a straight road increases its speed from 34.9 ft/s to 45.8 ft/s in a distance of 194 feet. If the acceleration is constant, how many seconds of time elapses while the motorcycle moves this distance?

Answer 1

The time elapsed while the motorcycle moves the distance is 4.8sec.

Given quantities are,

• Initial velocity=34.9 ft/s
• final velocity=45.8 ft/s
• distance travelled by motorcycle=194 ft

Since the acceleration of motorcycle is constant, we can apply two of the three equations of motion for uniform acceleration here.

• $v=u+a*t$ -first eq of motion under uniform acceleration.

• $v^{2} =u^{2} +2as$ -third eq of motion under uniform acceleration.

Substituting the given values in third eq of motion, we get

⇒$(45.8)^{2} =(34.9)^{2} +2*194*a$

this gives, acceleration, a=2.267 ft/$s^{2}$ (...1)

Now by substituting value of acceleration obtained above along with the given data in first eq of motion, we get;

⇒ 45.8=34.9+ 2.26*t

⇒45.8 - 34.9 = 2.26*t

⇒t = 10.9/2.26

⇒t=4.8 sec

Therefore, 4.8 sec have elapsed when the motorcycle has covered distance of 194 ft along with the given set of initial and final velocity.

To learn more about the motion under uniform acceleration, refer, brainly.com/question/25987127

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