# A car accelerates uniformly from rest to a speed of 21.0 km/h in 5.6 s. Find the distance the car travels during this time.Answer in units of m.

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The motion of the car has a constant acceleration. In this case we will use the expression of the acceleration and the expression of distance for the uniformly varied movement.

The acceleration is defined as:

a = (v – v0)/t

Where:

v – v0 → Variation of speed, at time t

The speed variation, is from zero to 21 km/h in a time of 5.6 seconds.

21 km/h = (21)(1000/3600) = 5.83 m/s

Then the acceleration is:

a = (5.83 – 0)/5.6 = 1.04 m/s^2

Now, we apply the expression of distance:

d = v0t + at^2/2

d = 0 + (1.04)(5.6)^2/2

d = 16.3 m

**Then, this car travels 16.3 m in a time of 5.6 s**

The car accelerates uniformly from rest to a speed of 21.0 km/h in 5.6 s.

The speed of 21 km/h can be converted to the units m/s by multiplying 21 with 1000/3600. This gives 35/6 m/s.

The acceleration of the car during the 5.6 seconds is (35/6)/(5.6) m/s^2

To determine the distance traveled by the car use the formula v^2 - u^2 = 2*a*s where u is the initial speed, v is the final speed, a is the acceleration and s is the distance traveled.

(35/6)^2 - 0^2 = 2*(35/6)/(5.6)*s

(35/6)^2 = 2*(35/6)/(5.6)*s

35/6 = 2/(5.6)*s

s = (35*5.6)/(6*2)

= 49/3 m

The distance traveled by the car is 49/3 m.

The time taken by the car to attain the final speed 21 kms/sec = 5.6 sec.

Sinceethe car started from the rest it initial speed , u = 0.

Therefore, the average speed of the car = (21+0)/2 =10.5 km/h= 2.916666 m/sec. So the car traverses with this average velocity 2.916666.. m/s for 5.6 seconds . So, it traverses 2.916666.*5.6 = 16.33 meters.

When an object travelling at an initial speed (velocity) of 'u' accelerates at the rate of 'a' the distance 's' travelled by it in time 't' is given by the formula:

s = [(u + v)/2]*t

it is given u = 21 km/h, v = 0, and t = 5.6

Therefore: s = [(u + v)/2]*t = [(21 + 0)/2]*5.6 = 58.8 m

Answer: Car travels a distance of 58.8 m.

The required equation of motion

v^2-u^2=2as (1), where v= final velocity and u is initial vlocity , a is the uniform acceleration and s is the displacement.

v= 21kms/hr = 21000/3600=5.833333m/s

But a =(v-u)/5.6 =(5.83333-0)/5.6 = 1.0416666m/s^2

v=5.833333m/s, u=0m/s, a=0.041666m/s^2 and s is to be found out.

By sustitutin in (1), we get:

5.83333^2-0^2 = 2*1.0416666*s

5.83333^2/(2*1.1416666) =s

s= 16.3333 m is the distance travelled by the car.