Question

Starting from a stationary position, Rahul paddles his bicycle to
attain a velocity of 6ms ^ - 1 30 s. Then he applies brakes such that the velocity of the bicycle comes down to 4 m * s ^ - 1 the next 5 s. Calculate the acceleration of the bicycle in both the cases.

Answer 1

Explanation:

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To answer these type of questions,

we have to introduce another physical quantity called acceleration, which is a measure of the change in the velocity of an object per unit time. That is,

$\boxed{\tt \purple{ acceleration = \frac{change \: in \: velocity}{time \: taken} } }$

If the velocity of an object changes from an initial value u to the final value v in time t, the acceleration a is,

$\boxed{ \tt \purple{ a = \frac{v - u}{t} }}$

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Solution:

In the first case:

$\tt \: Initial \: velocity, \: u = 0 \: \: \: \: \: \\ \\ \tt \: Final \: velocity, \: v = 6ms ^ {- 1 } \\ \\ \tt Time, t = 30 s \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

we know,

$\boxed{\tt \red{a = \frac{v - u}{t}}}$

Substituting the given values of u,v and t in the above equation, we get

$\tt a =\frac{ 6ms ^{ - 1 }- 0ms ^ {- 1}}{30s}\\ \\ \tt\red{ =0.2 m s^ 2}$

In the second case:

$\tt \: Initial \: velocity, \: u = 6ms ^{ - 1 } \\ \\ \tt Final \: velocity, \: v = 4ms ^ {- 1 } \: \: \\ \\ \tt Time, \: t = 5s \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Then,

$\tt a = \frac{4ms ^{ - 1} - 6ms ^{ - 1}}{5s}\\ \\ \tt\red{=-0.4 m s^ 2}$

The acceleration of the bicycle in the first case is .2ms² and in the second case, it is 4ms ^-2

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Answer 2

Answer:

- 0.4m/s²

Explanation:

Question:

Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of 6ms^-1

in 30s. Then he applies brakes such that the velocity of the bicycle cones down to 4ms^-1

in the next 5s. Calculate the acceleration of the bicycle in both the cases.

Solution :

For first case:

Initial velocity =0

Final velocity =60m/s

Time =30sec

We know that v = u + at

6=0+a×30

6=30a

a= 30/6 =0.2

a=0.2m/s²

For second case:

Initial velocity =6m/s

Final velocity =4m/s

Time taken =5sec

We know that v = u + at

4=6+a×5

4−6=5a

As we know that - and + are subtracted and sign comes of bigger no.

So , −2=5a

a=−2/5

a=−0.4m/s²

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