A car moving at constant speed of 30 m/s suddenly stalls at the bottom of a hill

Question

A car moving at constant speed of 30 m/s suddenly stalls at the bottom of a hill. The car undergoes a constant acceleration of -2m/s^2 while ascending the hill. Write equation for the position and the velocity as a function of time, taking x == 0 at the bottom of the hill where vo = 39 m/s. Determine the maximum distance traveled by the car up the hill after stalling. A student throws a vertically upward to another student in a window 4m above as shown in Figure 2.6. The keys are caught 1.5s later by the student. (a) With what initial velocity were the keys was thrown? (b) What was the velocity of keys just before they were caught?

Explanation / Answer

SOLUTION:

(A) from the equation of motion position of car can be given by

x= v*t + (1/2)*a*t 2

here v =30 m/s, a = -2 m/s2

so x = 30*t - (1/2)*2 *t2 = 30t-t2

(B) Equation of motion

Vfinal=Vinitial + a*t

before the stalling initial velocity of car was =30 m/s and acceleration was = -2 m/s2

when car reached to bottom of hill final velocity will be zero and it will give stalling time from the hill

so Vfinal =0

0= 30-2*t

t=15 s

again from equation of motion

x= v*t + (1/2)*a*t 2

Now we will calculate during this 15 second how much car travelled or displced

here v=15 because before reaching the bottom of the hill it have constant velocity of 30 m/s

x= 30*15 - (15)2

x=225 m

ans

NOTE:All these equation are valid becuase here acceleration and velocity are constant

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