http://ef.engr.utk.edu/c/sys/13/assess/paramimagev2.php?n=/www/c/ef151-2012-01/i

Question

http://ef.engr.utk.edu/c/sys/13/assess/paramimagev2.php?n=/www/c/ef151-2012-01/img/assess/bus-force-time-graph.png&p_F1=2710,158,79,&p_T1=8,160,233,0&p_F2=3530,309,59,&p_T2=14,300,233,0&p_MASS=2080,,,

A carbage bus' engine produces a net driving force whose magnitude varies as shown in the graph.
The 2080 kg bus is initially at rest.

A) Determine the acceleration of the bus at 8 seconds. (m/s2)
B) Determine the acceleration of the bus at 8 seconds. (mph/sec)
C) Determine the velocity of the bus at 8 seconds. (m/s)
D) Determine the velocity of the bus at 14 seconds. (m/s)

Explanation / Answer

given mass of the bus is 2080 kg

since force = mass * accleration

a) accleration at t=8 sec

accleration at 8 sec = force / mass = 2710/2080 = 1.303 m/sec2

b) accleration at t=14

accleration at 14 sec = 3530/2080 = 1.697 m/sec2

c) if in the above graph we divide force values by mass ( which is constant ) we will get accleration- time graph

the accleration time graph will be same as force-time graph in shape

now since we have the accleration-time graph

the area under the accleration-time graph gives the value of velocity

hence,

at t =8,

v = area of triangle under the graph from t=0 to t=8

v= .5* 8 * 1.303 = 5.212 m/sec                                      

d) velocity at t=14 is the area under the graph from t=0 to t=14

hence

v= area of triangle from t=0 to t=8 + area of trapezium from t=8 to t=14

v= .5 * 8 * 1.303 + .5 * 6 * (1.303 +1.697) ..............( area of trapezium with parellel sides along y axis)

v= 5.212 + 9 = 14.212 m/sec

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