Question

A car has an acceleration of \[8m{{s}^{-2}}\]. How much time is needed for it to reach a velocity of \[24m{{s}^{-1}}\] if it starts from rest. How far does it go during this period?

Answer 1

Hint: For solving such types of questions we have to use equations of motion according to the given data we can apply any of the three equations and get the results. Whenever it is mentioned in the question that the body starts from rest we will take initial velocity as zero. Whenever it is mentioned that brakes are applied we will take final velocity as zero.

Complete step-by-step solution:
Here it is given in the question that the body starts from rest. So the initial velocity of the car is taken as 0.
Let us assume the initial velocity of the car is\[u=0m{{s}^{-1}}\].
Since, here the acceleration which is given in the question is positive which means the car is accelerating, so velocity of car will increase and if acceleration given is negative then velocity of car is decreasing and finally the car comes to a stop.
Let us assume the positive acceleration of the car is represented as\[a=8m{{s}^{-2}}\].
When the car starts from rest and it accelerates then it takes some time to attain some velocity which is called as the final velocity of the car. So here the final velocity of the car is given which is 24ms-1.
Final velocity of the car is represented as\[v=24m{{s}^{-1}}\].
Suppose this final velocity of a car is acquired in\[t\sec \].
We have to calculate this time so we apply first equation of motion,
 \[v=u+at\]
Putting the values of all parameters in above equation we get,
\[\Rightarrow \]\[24=0+8t\]
\[\Rightarrow \]\[t=\dfrac{24}{8}\sec \]
\[\Rightarrow \]\[t=3\sec \]
So the car will take 3 seconds to attain this final velocity of\[24m{{s}^{-1}}\].
 Let us suppose in this time interval a car can travel through a displacement of S.
So we apply second equation of motion to calculate this displacement S,
\[s=ut+\dfrac{1}{2}a{{t}^{2}}\]
Putting the values of all parameters in above equation we get,
\[\Rightarrow \]\[s=(0)(3)+\dfrac{1}{2}(8){{(3)}^{2}}\]
\[\Rightarrow \]\[s=0+36\]
\[\therefore \]\[s=36m\]
So during this time interval the car has done a displacement of 36 m.
Finally the car will take a time period of 3secs to attain this velocity and displacement done in this time interval is 36m.

Note: When body falls under gravity then acceleration of body is changed to acceleration due to gravity. Acceleration due to gravity is represented by the symbol g. When the body falls towards earth then this acceleration due to gravity is taken as positive and when body going away from earth then this acceleration due to gravity is taken as negative.