Exercise 2.52 The acceleration of a bus is given by a x ( t )= t , where = 1.16

Question

Exercise 2.52

The acceleration of a bus is given by ax(t)=t, where = 1.16 m/s3 is a constant.

Part A

If the bus's velocity at time t1 = 1.09 s is 5.03 m/s , what is its velocity at time t2 = 2.20 s ?

2.22

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Part B

If the bus's position at time t1 = 1.09 s is 6.05 m , what is its position at time t2 = 2.20 s ?

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Exercise 2.52

The acceleration of a bus is given by ax(t)=t, where = 1.16 m/s3 is a constant.

Part A

If the bus's velocity at time t1 = 1.09 s is 5.03 m/s , what is its velocity at time t2 = 2.20 s ?

v =

2.22

  m/s  

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Incorrect; Try Again; 3 attempts remaining

Part B

If the bus's position at time t1 = 1.09 s is 6.05 m , what is its position at time t2 = 2.20 s ?

x =   m  

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Explanation / Answer

acceleration is given by:

a = alpha*t

we know that

a = dV/dt

dV/dt = 1.16*t

dV = 1.16*t*dt

integrating both sides

V = 1.16*t^2/2 + C

V = 0.58*t^2 + C

given that at t = 1.09, V = 5.03 m/sec

5.03 = 0.58*1.09^2 + C

C = 4.34

at t = 2.2 sec

V = 0.58*2.2^2 + 4.34 = 7.15 m/sec

So,

V = dx/dt = 0.58*t^2 + 4.34

dx = (0.58*t^2 + 4.34)*dt

x = 0.58*t^3/3 + 4.34*t + C

at t = 1.09 sec, x = 6.05 m

6.05 = 0.58*1.09^3/3 + 4.34*1.09 + C

C = 1.07

So at t = 2.2 sec

x = 0.58*2.2^3/3 + 4.34*2.2 + 1.07

x = 12.68 m

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