Question

17. A bicycle moves with a constant velocity of 5 km/h
for 10 minutes and then decelerates at the rate
1 km/h?, till it stops. Find the total distance covered
by the bicycle.
​

Answer 1

Given:

A bicycle moves with a constant velocity of 5 km/h  for 10 minutes and then decelerates at the rate  1 km/h till it stops.

To Find:

The total distance covered  by the bicycle

Solution:

We know that,

• $\orange{\underline{\boxed{\bf{Distance=Speed\times Time}}}}$
• Deacceleration is equal to acceleration in magnitude but have opposite sign with respect to acceleration
• According to third equation of motion for constant acceleration,

$\pink{\underline{\boxed{\bf{v^{2}-u^{2}=2as}}}}$

where,

v is final velocity

u is initial velocity

a is acceleration

s is displacement

$\rule{190}{1}$

We have two consider two cases:

Case-1: When bicycle is moving with constant speed

Here,

Speed of bicycle, s= 5 km/h

Time taken by bicycle,t= 10 min= 10/60h= 1/6 h

Let the distance covered by bicycle be d₁

So,

$\longrightarrow\rm{d_{1}=s\times t}$

$\longrightarrow\rm{d_{1}=5\times\dfrac{1}{6}}$

$\longrightarrow\rm{d_{1}=\dfrac{5}{6}\ km}$

$\rule{190}{1}$

Case-2: When bicycle is deaccelerating

Here,

Initial velocity of bicycle,u= 5 km/h

Final velocity of bicycle,v= 0 km/h

(Since, it stops finally)

Deacceleration of bicycle= 1 km/h

So,

Acceleration of bicycle,a= -1 km/h

Now, let the distance covered by bicycle be d₂

So, on applying third equation of motion on bicycle, we get

$\longrightarrow\rm{v^{2}-u^{2}=2as}$

$\longrightarrow\rm{(0)^{2}-(5)^{2}=2(-1)d_{2}}$

$\longrightarrow\rm{-25=-2d_{2}}$

$\longrightarrow\rm{-2d_{2}=-25}$

$\longrightarrow\rm{d_{2}=\dfrac{-25}{-2}}$

$\longrightarrow\rm{d_{2}=\dfrac{25}{2}\ km}$

$\rule{190}{1}$

Let the total distance covered  by the bicycle be D

So,

$\longrightarrow\rm{D=d_{1}+d_{2}}$

$\longrightarrow\rm{D=\dfrac{5}{6}+\dfrac{25}{2}}$

$\longrightarrow\rm{D=\dfrac{5+75}{6}}$

$\longrightarrow\rm{D=\dfrac{80}{6}}$

$\longrightarrow\rm\green{D=13.33\ km}$

Hence, the total distance covered by bicycle is 13.33 km.

Answer 2

$\sf{\red{\underline{\underline{\purple{Correct\:Answer\::{\dfrac{40}{3}}\:or\:13.33k.m}}}}}$

$\bigstar$ $\sf{\green{\underline{\underline{\orange{Given:-}}}}}$

• ${initial\:velocity\:(u)\:=\:5\:{\dfrac{k.m}{h}}}$
• Time (t) = 10 minutes = 1/6 hr.

☞ Also Given, the bicycle going on constant velocity .

• $\tt{\implies\:acceleration\:(a)\:=\:0}$

$\bigstar$ $\sf{\green{\underline{\underline{\orange{To\:Find:-}}}}}$

• The total distance covered by bicycle .

$\bigstar\sf{\green{\underline{\underline{\orange{CONCEPTS:-}}}}}$

(☞ Equation of motion when Acceleration is Constant,

• $\tt{v\:=\:u\:+\:at}$
• $\tt{s\:=\:ut\:+\:{\dfrac{1}{2}}\:at^2}$
• $\tt{v^2-u^2\:=\:2as}$

$\bigstar$ $\sf{\green{\underline{\underline{\orange{SOLUTION:-}}}}}$

☞ Here , acceleration (a) = 0

$\tt{s\:=\:ut\:+\:{\dfrac{1}{2}}\:at^2}$

$\tt{\implies\:s\:=ut}$

$\tt{\implies\:s\:=\:5\:\times\:{\dfrac{1}{6}}}$

$\tt{\implies\:s\:=\:{\dfrac{5}{6}}k.m}$ ----(1)

(☞ In case (2)

• u = 5 km/h
• v = 0
• a = 1 km/h²

(☞ Using Equation of Motion as,

• v² - u² = 2as
• => 0 - 5² = 2×(-1)×s
• => s = 25/2 k.m ------(2)

☞Hence total distance:Equation (1)+(2)

• $\tt{\implies\:13.33\:k.m}$

$\bigstar\:\underline{\boxed{\bf{\red{Required\:Answer\::13.33\:k.m\:}}}}$