Question

A car starting from Rest with uniform Acceleration acquires a velocity of 20 m/s in 20 sec. It travels at this velocity for
30 sec and is then bought to rest by being uniformly retarded, 150 sec after starting. Sketch A Velocity Time Graph And Calculate The

1.acceleration 2.retardation 3.total distance ​

Answer 1

Given :

✴ First case (A to B) :

▪ Initial velocity = zero

▪ Final velocity = 20mps

▪ Time interval = 20s

✴ Second case (B to C) :

▪ Constant velocity = 20mps

▪ Time interval = 30s

✴ Third case (C to D) :

▪ Initial velocity = 20mps

▪ Final velocity = zero

▪ Time interval = 150-(20+30) = 100s

To Find :

↗ Acceleration (A to B)

↗ Retardation (C to D)

↗ Total distance covered (A to D)

Concept :

➡ Acceleration is defined as the rate of change in velocity.

➡ It is a vector quantity.

➡ It can be positive/negative or zero.

➡ It has both magnitude as well as direction.

➡ SI unit : m/s²

Calculation :

⚽ Acceleration (A to B) :

$\red{\twoheadrightarrow}\tt\:a=\dfrac{v-u}{t_1}\\ \\ \red{\twoheadrightarrow}\tt\:a=\dfrac{20-0}{20}\\ \\ \red{\twoheadrightarrow}\underline{\boxed{\bold{\tt{\purple{a=1\:ms^{-2}}}}}}\:\orange{\bigstar}$

⚽ Retardation (C to D) :

$\green{\dashrightarrow}\tt\:a'=\dfrac{v'-v}{t_3}\\ \\ \green{\dashrightarrow}\tt\:a'=\dfrac{0-20}{100}\\ \\ \green{\dashrightarrow}\tt\:a'=-\dfrac{20}{100}\\ \\ \green{\dashrightarrow}\underline{\boxed{\bold{\tt{\tt{a'=-0.2\:ms^{-2}}}}}}\:\orange{\bigstar}$

[Note : -ve sign shows retardation.]

⚽ Total distance covered (A to D) :

$\pink{\Rightarrow}\tt\:d_{AD}=d_{AB}+d_{BC}+d_{CD}\\ \\ \pink{\Rightarrow}\tt\:d_{AD}=(ut_1+\dfrac{1}{2}a{t_1}^2)+(vt_2)+(vt_3+\dfrac{1}{2}a'{t_3}^2)\\ \\ \pink{\Rightarrow}\tt\:d_{AD}=(400+200)+(600)+(2000-1000)\\ \\ \pink{\Rightarrow}\tt\:d_{AD}=600+600+1000\\ \\ \pink{\Rightarrow}\underline{\boxed{\bold{\tt{\blue{d_{AD}=2200m=2.2km}}}}}\:\orange{\bigstar}$

Answer 2

Question will be easy when we divide it into three parts. i.e., A-B-C-D

Case - 1 :- ( A - B )

Initial velocity , u = 0 m/s

Final velocity , v = 20 m/s

Time , t = 20 sec

Case - 2 :- ( B - C )

Constant velocity , v = 20 m/s

Time , t = 30 sec

Case - 3:- ( C - D )

Initial velocity , u = 20 m/s

Final velocity , v = 0 m/s

Time , t = 150 - (20+30) = 100 sec

1. Acceleration : ( A-B )

Acceleration is defined as " Rate of change of velocity "

⇒ Acceleration , $\bold{a=\frac{(v-u)}{t} }$

⇒ a = (20-0)/20

⇒ a = 1 m/s²

2. Retardation : ( C-D )

Retardation is opposite to acceleration , having -ve sign extra.

⇒ a = (0-20)/100

⇒ a = -0.2 m/s²

Here -ve sign denotes retardation .

3. Total Distance ( A - D )

$\bold{D_{AD}=D_{AB}+D_{BC}+D_{CD}}\\\\\bold{D_{AD}=ut+\frac{1}{2}at^2 +vt+vt+\frac{1}{2}at^2}\\\\\bold{D_{AD}=200+600+2000}\\\\\bold{D_{AD}=2800\;m}\\\\\bold{\bf{\blue{D_{AD}=2.8\;km}}}$