A bike moving ahead on a straight highway passes a shop with a constant velocity of 18 m/s. After 5 seconds, it begins accelerating uniformly. 20 seconds after passing the shop, the bike's velocity is 63 m/s. Find the bike's speed 14 seconds after passing the shop
Answers
HELLO DEAR,
GIVEN:- A bike moving ahead on a straight highway passes a shop with a constant velocity of 18 m/s. After 5 seconds, it begins accelerating uniformly. 20 seconds after crossing the shop, the bike's velocity is 63 m/s.
TO FIND:- the bike's speed 14 seconds after crossing the shop.
SOLUTION:- speed of bike before crossing the shope V1 = 18 m/s.
speed of bike after crossing the shope V2= 63 m/s.
Time taken to cross the shope = 20 s.
So, acceleration of bike 'a',
V2 = V1 + at ( equation of motion)
a =( V2 - V1 ) / t
a = (63-18)/20
a = 45/20
a = 2.25 m/s²
To find the speed of bike after crossing the shope,
We take V1 = 63 m/s , a = 2.25 m/s², t = 14 s , V2= ?
V2 = V1 + at
V2 = 63 + 2.25 × 14
V2 = 63 + 31.5
V2 = 94.5 m/s.
Therefore speed of bike after crossing shope V2 = 94.5 m/s.
THANKS.
Given : A bike moving ahead on a straight highway passes a shop with a constant velocity of 18 m/s.
After 5 seconds, it begins accelerating uniformly
20 seconds after passing the shop, the bike's velocity is 63 m/s.
To Find : the bike's speed 14 seconds after passing the shop
Solution:
A bike moving ahead on a straight highway passes a shop with a constant velocity of 18 m/s.
Hence velocity also after 5 secs = 18 m/s
After 5 seconds, it begins accelerating uniformly = a
20 seconds after passing the shop, the bike's velocity is 63 m/s.
=> 20 - 5 = 15 secs
V = u + at
=> 63 = 18 + a(15)
=> 45 = 15a
=> a = 3 m/s²
bike's speed 14 seconds after passing the shop
14 - 5 = 9 secs
V = U + at
=> V = 18 + 3(9)
=> V = 45 m/s
bike's speed 14 seconds after passing the shop = 45 m/s
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