Question

Starting from stationery position Anil paddles his bicycle to attain a velocity of 10 m/s in 25 sec. Then he applies breaks such that he again comes to rest after next 5 secs. Calculate the acceleration of bicycle in both cases. Also, find the total distance covered.

Answer 1

Case 1: (In first half)
Step 1: Given

Initial velocity,$\text{u = 0 m/s}$

Final velocity,$\text{v = 10 m/s}$ v = 10 m/s

Time, t= 25 sec

Let a is acceleration, s be distance.

Step 2: Formula used

$\text{v = u+at}$ (First equation of motion)

$s=ut+\frac{1}{2}a{t}^2$ (Second equation of motion)

Step 3: Finding the acceleration

$10=0+a\left(25\right)$

$a=\frac{10}{25}=0.4m{s}^{-2}$

Step 4: Finding the distance

$s=ut+\frac{1}{2}a{t}^2$

$s=\left(0\right)t+\frac{1}{2}\left(0.4\right){\left(25\right)}^2=125m$

Hence, the distance covered is 125 m.

Case 2: (In second half)

Step 1: Given

Initial velocity, $\text{u= 10 m/s}$

Final velocity, $\text{v= 0 m/s}$

Time(t) = 5 sec

Step 2: Formula used

$\text{v = u + at}$ (First equation of motion)

$s=ut+\frac{1}{2}a{t}^2$(Second equation of motion)

Step 3: Finding the acceleration

$0=10+a\times 5$

$a=\frac{-10}{5}=-2m{s}^{-2}$

Step 4: Finding the distance

$s=ut+\frac{1}{2}a{t}^2$

$s=\left(10\right)\left(5\right)+\frac{1}{2}(-0.2){\left(5\right)}^2=47.5m$

Hence, the distance covered in this case is 47.5 m.

Now, total distance traveled by the Anil = 47.5 m + 125 m = 172.5 m