Question

a stunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp.
a. with why minimum speed must he drive off the horizontal ramp? the vertical heigh of th ramp is 1.5 m and the horizontal distance he must clear is 22m
b. if the ramp is now tilted upward, so that "takeoff angle" is 7° above the horizontal, what is the new minimum speed?

Answer 1

Answer:

a) Minimum speed must he drive off the horizontal ramp = 39.78 m/s

b) Minimum speed must he drive off the horizontal ramp with 7° above the horizontal  = 23.93 m/s

Explanation:

a) The height of ramp = 1.5 meter

   Horizontal distance he must clear = 22 meter

   The car is having horizontal motion and vertical motion. In case of vertical motion the acceleration on the car is acceleration due to gravity.

   We have equation of motion, $s= ut+\frac{1}{2} at^2$, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

 In case of vertical motion initial velocity = 0 m/s, acceleration = 9.8 $m/s^2$, we need to calculate time when displacement = 1.5 meter.

 $1.5=0*t+\frac{1}{2} *9.8*t^2\\ \\ t = 0.553 seconds$

So the car has to cover a distance of 22 meter in 2.119 seconds.

 So minimum speed required = 22/0.553 = 39.78 m/s

 Minimum speed must he drive off the horizontal ramp = 39.78 m/s

b) When the take of angle is 7⁰ the vertical speed of car is not zero = V sin 7 = 0.122 V

 So the in case of vertical motion we have initial velocity = 0.122 V, S = -1.5 meter( below ramp), acceleration = -9.8 $m/s^2$

Substituting

     $-1.5=0.122V*t-\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2-0.122Vt-1.5=0$

In case of horizontal motion

    Horizontal speed of car = V cos 7 = 0.993V

    So it has to travel 22 meter in t seconds

            0.993Vt = 22, Vt = 22.155 m

    Substituting in the equation $4.9t^2-0.122Vt-1.5=0$

    We will get $4.9t^2-0.122*22.155-1.5=0\\ \\ t = 0.926 seconds$

   Speed required = 22.155/0.926 = 23.93 m/s

  Minimum speed must he drive off the horizontal ramp with 7° above the horizontal  = 23.93 m/s

Answer 2

• The minimum speed must stunt driver drive off the horizontal ramp to jamb over 8 cars is 39.75 m/s.

• The new minimum speed, when the ramp is now tilted upward, so that "takeoff angle" is 7° above the horizontal is 23.93 m/s

What is velocity?

Velocity of a object is the ratio of distance traveled to the time taken.

Given information-

A stunt driver wants to make his car jump over 8 cars

All the cars parked side by side below a horizontal ramp.

• a.)The minimum speed must he drive off the horizontal ramp-

For part A of the problem we have given that,

The vertical height of the ramp is 1.5 m

The horizontal distance he must clear is 22 m

Put the values in the second equation of motion as,

$1.5=0\timest+\dfrac{1}{2}9.8\times t^2\\t=0.553\rm s$

As the speed is the ratio of distance covered to the time. Thus the minimum speed can be given as,

$v_{min}=\dfrac{22}{0.553} \\v_{min}=39.78\rm m/s$

The minimum speed must stunt driver drive off the horizontal ramp to jamb over 8 cars is 39.75 m/s.

• b.)  The new minimum speed, when the ramp is now tilted upward, so that "takeoff angle" is 7° above the horizontal-

For part A of the problem we have given that,

The takeoff angle is $7^o$ above the horizontal.

The equation can be written as,

$-1.5=vsin(7)\times t+\dfrac{1}{2}9.8\times t^2\\4.9t^2-0.122vt-1.5=0$

As the horizontal speed of the car is,

$v\cos (7)=0.993v$

Thus to travel 22 meters,

$0.993vt=22\\vt=22.155\rm s$

Put this value in the equation obtained we get,

$t=0.926\rm s$

Thus the minimum speed is,

$v_{min}=\dfrac{22.15}{0.926} \\v_{min}=23.93\rm m/s$

Hence The new minimum speed, when the ramp is now tilted upward, so that "takeoff angle" is 7° above the horizontal is 23.93 m/s

Hence,

• The minimum speed must stunt driver drive off the horizontal ramp to jamb over 8 cars is 39.75 m/s.
• The new minimum speed, when the ramp is now tilted upward, so that "takeoff angle" is 7° above the horizontal is 23.93 m/s

Learn more about the velocity here;

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