Question

A car starts from rest and moves with constant
acceleration of 4 m/s2 for 30 seconds. Then the
brakes are applied and the car comes to rest in
another 60 seconds. Find

Maximum velocity attained by car

Answer 1

Given :-

Initial velocity (u)=0

Acceleration (a)= 4m/s^2

Time (t)= 30 seconds

TO FIND :-

Maximum velocity of car

[ final velocity of car (v)]

Car will get it's highest (maximum) velocity at the end of 30 seconds as it will accelerate ,

after that the break applied and it started deceleration , and it's velocity will reduced.

So the we have to find the highest velocity of car at the end 30 seconds .

$\underline{\it{\dag \ \ \ By \ First \ Equation \ of \ motion }}$

$\bigstar{\boxed{\sf v=u+at}}$

Now put the given values in the formula

$\longmapsto\sf v=u+at\\ \\ \longmapsto\sf v= 0+4\times 30\\ \\ \longmapsto\sf v= 120m/s$

$\underline{\textsf{\textbf{\dag \ \ \ final \ velocity\ of \ car = 120m/s}}}$

$\underline{\red{\sf{Some \ important \ formula :-}}}$

$\sf \bullet v=u+at \\ \\ \sf\bullet s=ut+\dfrac{1}{2}at^2\\ \\\sf\bullet v^2=u^2+2as$

$\sf \dag \ \ \ where ,\\ \\ \sf u = Initial \ velocity\\ \\ \sf v= final \ velocity \\ \\ \sf s= Distance \\ \\ \sf a= acceleration$

Answer 2

Answer:

120 m/s

Explanation:

Given:

Initial velocity = u = 0m/s

acceleration = 4m/s²

Time = 30 seconds

To find:

Maximum velocity

The car will get maximum velocity at the end of 30 seconds as it will accelerate and after that as it deccelerates its velocity will be reduced , so the highest velocity is at the end of 30 seconds.

We can use the first equation of motion, which says:

V=u+at

Where,

V= final velocity (unknown)

u= initial velocity (0m/s)

a= acceleration (4m/s²)

t= time (30 seconds)

so, the final elocity will bethe maximum velocity attained by the car.

substituting the values, we get:

V=0+4×30

V=120 m/s

The maximum velocity is 120m/s