A student driver wants to make his car jump over eight cars parked side below a

Question

A student driver wants to make his car jump over eight cars parked side below a horizontal ramp. Part A) Whit what minimum speed must he drive off the horizontal ramp? The vertical height of the ramp is 1.5 m above the cars, and the horizontal distance he must clear is 20 m. Part B) If the ramp is now tilted upward, so that takeoff angle is 14 degree above the horizontal, what is the new minimum speed? A student driver wants to make his car jump over eight cars parked side below a horizontal ramp. Part A) Whit what minimum speed must he drive off the horizontal ramp? The vertical height of the ramp is 1.5 m above the cars, and the horizontal distance he must clear is 20 m. Part B) If the ramp is now tilted upward, so that takeoff angle is 14 degree above the horizontal, what is the new minimum speed? Part A) Whit what minimum speed must he drive off the horizontal ramp? The vertical height of the ramp is 1.5 m above the cars, and the horizontal distance he must clear is 20 m. Part B) If the ramp is now tilted upward, so that takeoff angle is 14 degree above the horizontal, what is the new minimum speed?

Explanation / Answer

a)

h=0.5gt2;

1.5 = (0.5)(10)t2;

t=0.548 sec;

let u be the minimum speed:

then distance travelled by car in this time should be equal to 20 meters i.e.

ut = 20;

u= 20/t;

=> u= 20/0.548;

u= 36.496 m/s

b)

let the minimum speed in this case be u.

u(cos14)t = 20; --------------------------------equation 1

-1.5 = u(sin14)t - 0.5gt2; --------------------equation 2

From equation 1:

ut = 20/cos14;

ut = 20.612;

Substitute ut in equation 2:

-1.5 = 4.987 - 5t2;

=> t = 1.139 sec

Substitute t in equation 1:

u = 20/(1.139)(0.97);

=> u = 18.102m/s

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