Question

Starting from a stationary position, Seetha paddles her bicycle to attain a velocity of 6 m/s in 30s. Then she applies brakes such that the velocity of the bicycle comes down to 4 m/s in the next 5s. Calculate the acceleration of the bicycle in both the cases.

• A. ,
• B. 2 , 3
• C. , -2.5
• D. 3 , 4

Answer 1

Step 1: Given that:

In the first case, when Seetha starts her motion;

The initial velocity($u_1$) of Seetha = $0$

The final velocity($v_1$) of Seetha = $6m{s}^{-1}$

Time taken ($t_1$) = $30s$

In the second case, when Seetha applies brakes;

The initial velocity($u_2$) = The final velocity in the first case

= $6m{s}^{-1}$

Final velocity($v_2$) = $4m{s}^{-1}$

Time taken($t_2$) = $5sec$

Step 2: Calculation of acceleration in the first case:

Using the first equation of motion; $v=u+at$

Where u= Initial velocity, v= final velocity, t= time and a= acceleration in the body

$6m{s}^{-1}=0+a\times 30s$

$\Rightarrow 30a=6$

$\Rightarrow a=\frac{6}{30}$

$\Rightarrow a=0.2m{s}^{-2}$

$\Rightarrow a=\frac{1}{5}m{s}^{-2}$

Step 3: Calculation of acceleration in the second case:

Again using the first equation of motion, we get;

$4m{s}^{-1}=6m{s}^{-1}+a\times 5sec$

$\Rightarrow 6+5a=4$

$\Rightarrow 5a=-2$

$\Rightarrow a=\frac{-2}{5}$

$\Rightarrow a=-0.4m{s}^{-2}$

$\Rightarrow a=-\frac{2}{5}m{s}^{-2}$

(Negative sign shows that it is retardation.

Thus,

Acceleration in the first case = $\frac{1}{5}m{s}^{-2}$

Acceleration in the second case = $-\frac{2}{5}m{s}^{-2}$

So correct option is (a)$\frac{1}{5}m{s}^{-2}$ ,$-\frac{2}{5}m{s}^{-2}$