28 PE
Question
A powerful motorcycle can accelerate from rest to 26.8 m/s (100 km/ h) in only 3.90 s. (a) What is its average acceleration? (b) How far does it travel in that time?
Step-by-Step Solution
Verified(a) The average acceleration of motorcycle is \({\bf{6}}{\bf{.9}}\;{\bf{m/}}{{\bf{s}}^{\bf{2}}}\).
(b) The distance travelled by motorcycle is \({\bf{52}}\;{\bf{m}}\).
(a)
Given Data:
The initial velocity of motorcycle is \(u = 0\)
The final velocity of motorcycle is \(v = 26.8\;{\rm{m}}/{\rm{s}}\)
The deceleration of racer is \(a = 2\;{\rm{m}}/{{\rm{s}}^2}\)
The time for distance travelled is \(t = 3.90\;{\rm{s}}\)
The average acceleration of the motorcycle is found by using first equation of motion and distance travelled by third equation of motion.
The average acceleration of motorcycle is given as
\(v = u + at\)
Here, \(a\) is the average acceleration of motorcycle.
Substitute all the values in the above equation.
\(\begin{array}{c}26.8\;{\rm{m}}/{\rm{s}} = 0 + a\left( {3.90\;{\rm{s}}} \right)\\a = 6.9\;{\rm{m}}/{{\rm{s}}^2}\end{array}\)
Therefore, the average acceleration of motorcycle is \(6.9\;{\rm{m}}/{{\rm{s}}^2}\).
(b)
The distance travelled by motorcycle is given as
\({v^2} = {u^2} + 2ad\)
Substitute all the values in the above equation.
\(\begin{array}{c}{\left( {26.8\;{\rm{m}}/{\rm{s}}} \right)^2} = {\left( 0 \right)^2} + 2\left( {6.9\;{\rm{m}}/{{\rm{s}}^2}} \right)d\\d = 52\;{\rm{m}}\end{array}\)
Therefore, the distance travelled by motorcycle is \(52\;{\rm{m}}\).