A driver realizes too late that the 10 degree hill he is driving up levels off a

Question

A driver realizes too late that the 10 degree hill he is driving up levels off abruptly as shown in the diagram. At the instant he slams on his brakes his velocity is 25 m/s and he is 15 meters from the end of the hill. The coefficient of friction between his tires and the road is 0.6. Find the speed of the car at the top of the hill. Find the maximum height of the car above the level road. Suppose that the driver had not slowed down, and had reached the level art of the road at a speed of 25 meters per second. Find the horizontal distance that the car would travel while airborne.

Explanation / Answer

Here ,

i)

deceleration of the car = g*sin(10) + uk*g*cos(10)

a = 9.8 (sin(10) + 0.6 * cos(10))

a = 7.34 m/s^2

Now, using third equaiton of motion

v^2 - u^2 = 2*a*d

v^2 - 25^2 = -2*7.34 * 15

v =20.12 m/s

his speed at the top of hill is 20.12 m/s

ii)

Now, for theta = 10 degree

Hmax = (v*sin(10))^2/2g

Hmax = (20.12*sin(10))^2/(2*9.8)

Hmax = 0.505 m

the maximum height of the car is 0.505 m

iii)

for v = 25 m/s

horizontal distance , d = (25*sin(2*10))^2/9.8

d = 6.1 m

the distance is 6.1 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.