the speedometer of bike reads 52km/h at 8.00 hours and after travelling for a certain distence it reads 66km/h at8.25 hours.find the distance travelled during this time.
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3
distance traveled is avg speed * time duration
= (52+66)/2 kmph * (25/60) hrs = 59 * 5/ 12 = 24.58 km
= (52+66)/2 kmph * (25/60) hrs = 59 * 5/ 12 = 24.58 km
Answered by
0
assuming uniform acceleration
v=u+at
a= (v-u)/t =(66-52)/0.25 =56
s= ut +0.5 at² = 52 x 0.25 + 0.5 x 56 x 0.25²
= 0.25( 52+ 7)
=59/4
=14.75 km
v=u+at
a= (v-u)/t =(66-52)/0.25 =56
s= ut +0.5 at² = 52 x 0.25 + 0.5 x 56 x 0.25²
= 0.25( 52+ 7)
=59/4
=14.75 km
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