Question
Question: A car starting from rest has a speed of \(30\,kmh{r^{ - 1}}\) at any one instant. Two seconds later ...
A car starting from rest has a speed of 30kmhr−1 at any one instant. Two seconds later its speed is 36kmhr−1 and two seconds after that it is 42kmhr−1.What is its acceleration in ms−2?
Solution
This problem can be solved using the first equation of motion. First convert the given speeds and calculate acceleration using the equation of motion. The car starts from rest, hence its initial velocity will be 0. After two seconds the velocity is given. Also, the thing to do first is to convert all the given quantities in standard SI units.
Formula Used:
To solve the problem we have used the following formula,
v=u+at v=u+at
Where v represents the final velocity, u represents the initial velocity, a represents acceleration and t represents time taken.
Complete step by step answer:
In the question it is given that when the car starts from rest its speed is 30kmhr−1 which means its initial velocity is 30kmhr−1. Two seconds later its speed is 36kmhr−1and two seconds after that it is 42kmhr−1. So, we have to first calculate the acceleration in ms−2. Firstly, we will have to convert the speeds from kmhr−1 toms−1. So, converting each of the given velocities we can write,
u1=30kmhr−1 ⇒u1=360030×1000ms−1 ⇒u1=36300ms−1 ⇒u1=325ms−1
⇒u2=36kmhr−1 ⇒u2=360036×1000ms−1 ⇒u2=36360ms−1 ⇒u2=10ms−1
⇒u3=42kmhr−1 ⇒u3=360042×1000ms−1 ⇒u3=36420ms−1 ⇒u3=335ms−1
Now, for calculating acceleration from the formula we use the values provided in the question as,
u=u1=325ms−1
⇒v=u2=10ms−1
⇒t=2s........................(given in the question)
Therefore, substituting the respective values of u,v,t in the formula we get,
v=u+at
⇒10=325+2a
⇒2a=10−325
⇒2a=330−25 ⇒2a=35
⇒a=3×25 ∴a=65ms−2
Hence, acceleration is 65ms−2.
Note: This problem requires conversion of units many times. So, the conversions must be done carefully so that all the quantities are in the same system of units. Also, it is important to understand which of the given velocity represents initial velocity and which one represents final velocity. While doing problems involving Newton’s equations of motion, we have to keep in mind the sign conventions. If a body is accelerating then its acceleration is positive and if the body is coming to stop after some time then it is accelerating and its acceleration is negative.