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A car that is initially moving at 7.50 m/s begins to accelerate forward uniformly at 0.550 m/s2.

1 Answer

1 vote

Answer:

Step-by-step explanation:

How long after beginning to accelerate does it take the car to move 3.5km?

Vi = car initial velocity

a = increase in velocity of the car per second

s = distance travelled by the accelerating car in time t

VF = velocity of the accelerating car at time t

Given from question

Vi = 7.5 m/s

a = 0.55 m/s²

s = 3.5km

t = ?

S = Vi * t + 1/2 * acceleration * time²

3500m = 7.5m/s * t + 1/2 * 0.55 * t²

0.55t² + 15t - 7000 = 0

We will use roots of a quadratic equation to find t

ax² + bx + c = 0

x = (-b ± (√b² - 4*ac)) / 2a

t = (-15 ± (√15² + 4 * 0.55 * 7000)) / 2 * 0.55

t = -15 ± 125 / 1.1

t = -140 / 1.1 ║t = 110 / 1.1

t = -127.27 ; 100

So, the equation of 0.55t² + 15t - 7000 = 0 is satisfied with two values of t = -127.27 & 100

Since, we are talking about time, we will eliminate minus value and we will use the positive value.

Therefore, the time needs to accelerate to move the car by 3.5km after the beginning is 100 seconds.

Question B

Given from question

S = 3.5km = 3500m

Velocity Final = ?

Velocity Final² = Vi² + 2*acceleration*distance (s)

Velocity Final² = 7.5² + 2 * 0.55 * 3500

Velocity Final² = 3,906.25

Velocity Final = 62.5m/s

Since we already found the accelerating car travelled 3500m in 100s. we could also use the other formula :

Velocity Final = Vi + acceleration * time

= 7.5 + (0.55*100)

= 62.5 m/s

Therefore, the car is moving 62.5m/s after traveled 3.50km.

User Andrew Bannerman
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