Question

a car acceleration form 15 km / h to 36 km/h in 5 sec calculate and calculate 1) the acceleration and 2) distance cover by car in this time.
place help me​

Answer 1

Answer:

21 km /h is acceleration

7.2 × 21 is the time taken

Answer 2

GIVEN:

• Initial Velocity (u) of car = 15 km/hr
• Final velocity (v) of car = 36 km/hr
• Time taken (t) by the car = 5 sec

TO FIND:

• What is the acceleration and Distance covered by the car ?

SOLUTION:

We have given the initial and final velocity of car are 15 km/hr and 36 km/hr respectively.

Change the velocity in m/s:-

◉ Initial velocity = 15 km/hr

◉ u = 15 $\times$ $\sf{\dfrac{5}{18}}$

◉ u = 4.16 m/s

Final velocity = 36 km/hr

◉ v = 36 $\times$ $\sf{\dfrac{5}{18}}$

◉ v = 10 m/s

Let the acceleration of the car be 'a' m/s²

To find the acceleration of the car, we use the first Equation of motion:-

❰ v = u + at ❱

According to question:-

➸ 10 = 4.16 + a $\times$ 5

➸ 10 –4.16 = 5a

➸ 5.84 = 5a

➸ $\sf{\cancel\dfrac{5.84}{5}}$ = a

❮ a = 1.168 m/s² ❯

To find the distance travelled by the car, we use the formula:-

❰ s = ut + $\bf{\dfrac{1}{2}}$ at²❱

According to question:-

➸ s = 4.16 $\times$ 5 + $\sf{\dfrac{1}{2}}$ $\times$ 1.168 $\times$ (5)²

➸ s = 20.8 + 0.584 $\times$ 25

➸ s = 20.8 + 14.6

❮ s = 35.4 m ❯

❝ Hence, the acceleration and Distance covered by the car is 1.168 m/s² and 35.4 m respectively. ❞

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