6) Uniformlyacceleratedmotion Time oftravel V The initial velocity of an object

Question

6) Uniformlyacceleratedmotion Time oftravel V The initial velocity of an object is 5 m/s. The object accelerates at the rate of does it take the object to reach the velocity of 35 m/s?o5 A) 15 s 3 m/s?. How long B) 10 s 7) Uniformly accelerated motion, Displacement C) 2 s 3S 5 A car starts from rest and accelerates at the rate of 6 m/s2 Find the displacement at t-5 s? A) 75 m B) 114 m C) 30 m D) 15 m 8) Uniformly accelerated motion Stopping distance A car begins to slow down when its velocity is 12 m/s. It travels 120 m before coming to a stop What is the acceleration of the car (the acceleration is negative)? A)-10 m/s2 B)-0.6 m/s2 C) -4 m/s2 D)-5 m/s2 9) Free fall. Time of fall A ball is released from rest from the height of 14 m. How long does it take to fall to the ground? A) 1.2 s B) 0.8 s D) 2.2s 10) Free fall. Height A ball is thrown downward from the height h with an initial speed of 25 m/s. It takes 2 s to fall to the ground. Find the height h. A) 70 m B) 30 m C) 50 m D) 20 m 11) Free fall. Maximum height A ball is thrown upward with velocity of 15 m/s. Find the maximum height reached by the ball. A) 26.4 m B) 7.3 m C) 18.1 m 11.5 m 12) Free fall. Time to reach the maximum height A ball is thrown upward with velocity of 15 m/s. Find the time it takes to reach the maximum height A) 0.9 s C) 2.7 s D)4.1s B) 1.5 m 13) Free fall. Total time of flight A ball is thrown upward with velocity of 15 m/s. Find the time it takes to fall back to the ground. 6) 3.1 s D) 8.2s C) 2.6 s Slope-intercept equation for a straight line y=mx + b B) 5.8 s

Explanation / Answer

Solution:

Let us go to the basics first.

Answer (7): From Newton's Equation of motion:

S = ut + (1/2)at2

=> S = 0 + (1/2)*6*52 [u = 0 since rest; t = 5 s; a= 6 m/s2]

=> S = 0 + (1/2)*6*52

=> S = 75 m (Answer 7===>option A)

Answer (8): From Newton's Equation of motion:

v2 = u2 + 2as [v = 0 on stopping; u = 12 m/s; s = 120 m]

=> 02 = 122 + 2a*120

=> a = - 0.6 m/s2 (Answer 8 ===> Option B)

Answer (10):

From Newton's Equation of motion:

S = ut + (1/2)at2 [u = 25 m/s; t = 2 s; s = h; a = 10 m/s2]

=> S = 25*2+ (1/2)*10*22

=> S = 70 m (Answer 10 ====> Option A)

Answer (12):

At maximum height, v = 0.

From Newton's equation of motion:

v = u + at

=>0 = 15 - 10t

=> t = 1.5 s (Answer 12 ===> Option B)

Thanks!!!

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