Question

a toy car powered by water and compressed air when full charged and released it accelerates at 2m/s² for 3 sec and then it begans to slow down until it comes to rest.
it covers a combined distance of 27 m find its retardation speed and time of motion.​

Answer 1

Correct Question

A toy car powered by water and compressed air when full charged and released it accelerates at 2 m/s² for 3 sec and then it began to slow down until it comes to rest.  It covers a combined distance of 27 m. Find its retardation and time of motion.​

Given:

A toy car accelerates at 2 m/s² for 3 sec and then it began to slow down until it comes to rest.

Total distance covered by car,D= 27m

To Find:

Retardation and time of motion

Solution:

We know that,

• Retardation is equal to acceleration in magnitude and opposite to acceleration in direction or sign

• According to first equation of motion for constant acceleration,

$\purple{\boxed{\bf{v=u+at}}}$

• According to third equation of motion for constant acceleration,

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

where,

v is final velocity

u is initial velocity

a is acceleration

s is displacement

t is time taken

$\rule{190}{1}$

We have to consider two cases:

Case-1: When toy car is accelerating

Here,

Initial velocity of toy car,u= 0 m/s

(Since, it starts from rest)

Acceleration of toy car,a= 2 m/s²

Time taken by toy car,t= 3 s

Now,

Let the final velocity of toy car be v and distance covered by toy car be s

So, on applying first equation of motion on toy car, we get

$\longrightarrow\rm{v=u+at}$

$\longrightarrow\rm{v=0+2(3)}$

$\longrightarrow\rm{v=6\ m/s}$

Also, on applying third equation of motion on toy car, we get

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

$\longrightarrow\rm{(6)^{2}-(0)^{2}=2(2)s}$

$\longrightarrow\rm{36=4s}$

$\longrightarrow\rm{4s=36}$

$\longrightarrow\rm{s=\dfrac{36}{4}}$

$\longrightarrow\rm{s=9\ m}$

Case-2: When toy car is retarding

Here,

Initial velocity of toy car,u'= 6 m/s

Final velocity of toy car,v'= 0 m/s

(Since, toy car stops finally)

Let the displacement of toy car for second case be s'

It is also given that total distance covered by car is 27 m

So, according to the question

$\longrightarrow\rm{D=s+s'}$

$\longrightarrow\rm{27=9+s'}$

$\longrightarrow\rm{27-9=s'}$

$\longrightarrow\rm{18=s'}$

$\longrightarrow\rm{s'=18\ m}$

Let the acceleration of toy car be a' and t' be the time taken by toy car

So, on applying third equation of motion on toy car, we get

$\longrightarrow\rm{(v')^{2}-(u')^{2}=2a's'}$

$\longrightarrow\rm{(0)^{2}-(6)^{2}=2a'(18)}$

$\longrightarrow\rm{-36=36a'}$

$\longrightarrow\rm{36a'=-36}$

$\longrightarrow\rm{a'=\dfrac{-36}{36}}$

$\longrightarrow\rm{a'=-1\ m/s^{2}}$

Since, acceleration is -1 m/s²

Therefore, retardation is 1 m/s²

Also, on applying first equation of motion on toy car, we get

$\longrightarrow\rm{v'=u'+a't'}$

$\longrightarrow\rm{0=6+(-1)t'}$

$\longrightarrow\rm{-6=-t'}$

$\longrightarrow\rm{-t'=-6}$

$\longrightarrow\rm{t'=6\ s}$

Let the total time of motion be T

So,

$\longrightarrow\rm{T=t+t'}$

$\longrightarrow\rm{T=3+6}$

$\longrightarrow\rm\green{T=9\ s}$

Hence, retardation of toy car is 1 m/s² and total time of motion is 9 s.