a bus starts from rest and attain a velocity of 45m/s after 10 seconds. find the acceleration and the distance travelled in 5 sec

2 thoughts on “a bus starts from rest and attain a velocity of 45m/s after 10 seconds. find the acceleration and the distance travelled in 5 sec”

1. Given:

   u=0m/s because bus is starting from rest

   v=45m/s bus attains this speed

   t=10s at this duration of time

   To Find: a=? and s=? at t=5s and t=10s

   So, let t=10s

   Now,

   v’= 45/2= 22.5m/s because at 10s it attains 45 so in 5s it will attain half of 45

   So,

   we have the formula for acceleration,

   a=(v+u)/t in this case u is 0

   so, a=v’/t v’ we took because to find for 5s

   now substituting there values, we get,

   a=22.5/5

   a=4.5m/s²

   Now,

   for t=10s

   so, same formula we use instead of v’ now we use v which is final velocity

   So,

   a=45/10

   a=4.5m/s²

   Therefore the acceleration at t=5s is 4.5m/s² and at t=10s is 4.5m/s²

   therefore we can also conclude that the bus is uniformly accelerated till time t=10s

   Now, we have to find distance s=?

   so for t=5s

   we have,

   s=ut+(1/2)at²

   s=0*5+(1/2)*4.5*(5)²

   s= 0+(1/2)*112.5

   s=56.25m

   And,

   for t=10s

   s=ut+(1/2)at²

   s=0*10+(1/2)4.5*(10)²

   s=(1/2)*450

   s=225m

   There for the distance travelled at t=5s is 56.25m and at t=10s is 225m

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2. Answer:

   225,450

   Explanation:

   after 5 min it will be 45×5 =225

   after 10 min it will be 45×10 = 450

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