Question

A bus starts from rest and attains a velocity of 45 m/s after 10seconds .Find acceleration and distance travelled in 5seconds and in 10 seconds .

Answer 1

According to question ,

u = 0 m/s

v = 45 m/s

t = 10 sec

a = (v+u)/t

a = (45+0)/10

a = 45/10

a = 4.5 m/s^2. (Ans. 1)

Now , according to question ,

t = 5 sec

Applying second equation of motion , we get ,

s = ut + (1/2)at^2

s = 0x5 + (1/2) 4.5x(5)^2

s = (1/2) 4.5x25

s = 1/2 x 112.5

s = 56.25 m. (Ans. 2)

Now , According to question ,

t = 10 sec

Applying second equation of motion , we get ,

s = 0x10 + (1/2) 4.5x(10)^2

s = (1/2) 4.5x100

s = (1/2) 450

s = 225 m. (Ans. 3)

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Answer 2

Explanation:

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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