Question

A car accelerates from rest at a constant rate of 2 m/s2 for some time. Then, it retards at a constant rate of 4 m/s2 and comes to rest. If it remains in motion for 3 seconds, then the maximum speed attained by the car is

Answer 1

Explanation:

Let v,t

1

,t

2

be the maximum velocity attained, time for which car accelerated and the time for which car retarded respectively

⇒t

1

=(v−u)/a=(v−0)/2=v/2

⇒t

2

=(v−u)/a=(0−v)/(−4)=v/4

Now it is given that,

⇒t

1

+t

2

=3

⇒v/2+v/4=3

⇒v=4m/s

Hence correct answer .

Answer 2

$\huge\red{\underline{{\boxed{\textbf{QUESTION}}}}}$

A car accelerates from rest at a constant rate of 2 m/s2 for some time. Then, it retards at a constant rate of 4 m/s2 and comes to rest. If it remains in motion for 3 seconds, then the maximum speed attained by the car is

$\huge\red{\underline{{\boxed{\textbf{ANSWER}}}}}$

Let car gain a maximum velocity =v m\s

And let for getting this highest speed car spend t time

Now remaining time is 3-t sec

Given in question acceleration is 2m\s2

Now by the first equestion of motion

v=u+at. ( v =final velocity,t=time,u=initial velocity ,a=acceleration);

v=0+2t. ___1. (Because initial velocity is zero)

Now again given

Retardation 4m\s2

v=u+at;

0=v+(-4)(3-t). ______2

On putting the value of v in equestion 2

0=2t-12+4t

Hence t=2sec

After 2 sec car will get maximum velocity,

Now on putting value of t in equestion 1

v=2×2

v=4 m\s

@HarshPratapSingh