

InterviewSolution
Saved Bookmarks
1. |
A bus is travelling at speed of 15m/s. Then it accelerates for 5s at a rate of 10m/s2. Find the final velocity and distance covered by the bus during this process. |
Answer» <html><body><p>-The speed of the bus = 15m/sThe acceleration of bus = 10m/s²Time <a href="https://interviewquestions.tuteehub.com/tag/taken-659096" style="font-weight:bold;" target="_blank" title="Click to know more about TAKEN">TAKEN</a> by the bus = 5sTo Find :-The <a href="https://interviewquestions.tuteehub.com/tag/final-461168" style="font-weight:bold;" target="_blank" title="Click to know more about FINAL">FINAL</a> velocity of bus (v) = ?The distance covered by the bus (s) = ?Solution :-To calculate the final velocity of bus and distance travelled by the bus , at first we have to use first equation of motion to calculate final velocity of the bus then calculate distance travelled by the bus by applying <a href="https://interviewquestions.tuteehub.com/tag/2ns-1838421" style="font-weight:bold;" target="_blank" title="Click to know more about 2NS">2NS</a> equation of motion.Calculation for Final velocity :-V = ? U = 15m/s. T = 5s. A = 10m/s² :-⇢ First equation of motion :-⇢ V = U + A × T⇢ V = <a href="https://interviewquestions.tuteehub.com/tag/15-274069" style="font-weight:bold;" target="_blank" title="Click to know more about 15">15</a> + 10 × 5⇢ V = 15 + 50⇢ V = 65m/sTherefore, final velocity of the bus (v) = 65m/sCalculation for Distance covered by bus :-S = ? U = 15m/s. T= 5s. A = 10m/s²⇢ 2nd equation of motion :-⇢ S = UT + 1/2 × A × T²⇢ S = 15 × 5 + 1/2 × 10 × 5²⇢ S = 75 + 5 × 25⇢ S = 75 + 125⇢ S = 200mTherefore, distance covered by the bus (s) = 200m</p></body></html> | |