“Rate of change of velocity of an object with respect to time is called acceleration.”
acceleration = change in velocity/ time taken
\( a= \frac{v}{t}\)
a = (vf – vi)/t
Following are the three types of acceleration
“If the velocity of a body increases continuously then the acceleration is said to be positive.”
“If the velocity of a body decreases continuously then the acceleration is said to be negative is called deceleration or retardation.”
“A constant rate of change of velocity is called uniform acceleration.”
\(a = \frac{v_f – v_i}{t}\)
“A car travelling at 30 m/s accelerates at a constant rate of 2 m/s2 for 10 seconds.
Calculate its final velocity and the distance travelled during this time.
Solution
Step 1: Write the known quantities and point out quantities to be found.
Vi = 30 m/s
a = 2 m/s^2
t = 10 s
Vf = ?
distance = ?
Step 2: Write the formula and rearrange if necessary.
Vf = Vi + at
distance = Vi*t + (1/2)at2
Step 3: Put the value in the formula and calculate.
Vf = 30 + 210 = 50 m/s
distance = 3010 + (1/2)2(102) = 350 m
The final velocity of the car is 50 m/s and it travels a distance of 350 m during the 10 seconds of acceleration.”