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:

Given

Initial velocity=18 km/hr = 5 m/s

Final velocity=v=36km/h=10 m/s

Time taken = 5 seconds

We need to calculate acceleration and distance traveled

We know that

acceleration a = (v-u)/t

Substituting the given values in the above equation we get,

a=(10-5)/5 =5/5= 1 m/s2

We know that distance travelled is calculated by the formula

Distance travelled S = u t + (1/2) a t2

Substituting the given values in the above equation we get,

S =5×5 + (1/2)×1×52

25 + (1/2) X 25

25 + 12.5

= 37.5 m

Hence

acceleration a =1 m/s2

Distance travelled S =37.5 m

Answer 2

Answer:

• Initial Velocity (u) = 15 km/hr = 4.16 m/s
• Final Velocity (v) = 36 km/hr = 10 m/s
• Time taken (t) = 5 seconds

★ According to Question :

By using first equation of motion we get :

→ v = u + at

→ 10 = 4.16 + a(5)

→ 10 - 4.16 = 5a

→ 5.84 = 5a

Dividing both the sides by 5 we get :

→ a = 1.168 m/s²

Now, let's find the distance travelled by the car

By Using Third Equation of motion we get :

⇢s = ut + ½at²

⇢s = (4.16)(5) + ½ (1.168)(5)²

⇢s = 20.8 + 0.584 × 25

⇢s = 20.8 + 14.6

⇢s = 35.4 m

Therefore,

• Acceleration produced by the car is = 1.168 m/s²
• Distance travelled by the car = 35.4 m