Question

the velocity of a car increases from 36 km/h to 108 km/h in 10s. find its acceleration ​

Answer 1

Answer:

Given :

➡️ u = initial velocity = 36km/hr

that is, u = 36×1000/3600 m/s

that is, u = 10m/s

➡️v= final velocity = 108km/hr

that is, v= 108×1000/3600m/s

that is, v= 30m/s

➡️ time = t = 10s

To find :

➡️ acceleration of the car

Solution :

we know that

acceleration(a) = (v-u)/t

that is, a = (30-10)/10m/s square

that is, a = 20/10m/s square

that is , a = 2 m/s square

Hence, acceleration of the car is 2 m/s square.

Answer 2

Answer:

The velocity of a car increases from 36 km/h to 108 km/h in 10s. Its acceleration will be  ​$2 \ m/s^{2}$

Explanation:

Acceleration of the car can be defined as the rate of change of velocity with respect to time.

If u is the initial velocity, v is the final velocity and time is t,

then the acceleration (a) is given by,

$a = \frac{v-u}{t}$     ...(1)

From the question, it is given that,

u   =  $36 \ km/h$  $= 36\times \frac{5}{18}$  $= 10\ m/s$

v    = $108\ km/hr = 108\times \frac{5}{18} = 30\ m/s$

t     =  $10\ s$

Equation(1) becomes,

$a = \frac{30-10}{10}$

    $= 2 \ m/s^{2}$

Thus the acceleration of the car  $= 2 \ m/s^{2}$