A bus moving with a speed of 54 km/h. On applying brakes, it stopped in 6 seconds. Calculate the acceleration and the distance travelled before stopping.
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A bus moving with a speed of 54 km/h. On applying brakes, it stopped in 6 seconds. Calculate the acceleration and the distance travelled before stopping.
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Answer:
54×1000/3600=15m/s
a=(Vt-Vo)/t
a=(0–15)/8
a=-15/8=-1.875m/s^2 or -1.875m/s/s
Distance=(Vo+Vt)×t/2
Distance=15×8/2=60m
Explanation:
This is how I would do it:
First, convert 54km/h to m/s. This equals 15m/s.
Secondly, I assume that the bus breaks with constant acceleration. When this is true, theres a kinematic equation that says:
v=v0⋅a⋅t" role="presentation" style="display: inline-table; font-style: normal; font-weight: normal; line-height: normal; font-size: 15px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">v=v0⋅a⋅tv=v0⋅a⋅t
Which can be rewritten as:
a=v−v0t" role="presentation" style="display: inline-table; font-style: normal; font-weight: normal; line-height: normal; font-size: 15px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">a=v−v0ta=v−v0t
And we know the following values:
v=0" role="presentation" style="display: inline-table; font-style: normal; font-weight: normal; line-height: normal; font-size: 15px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">v=0v=0
v0=15m/s" role="presentation" style="display: inline-table; font-style: normal; font-weight: normal; line-height: normal; font-size: 15px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">v0=15m/sv0=15m/s
t=8s" role="presentation" style="display: inline-table; font-style: normal; font-weight: normal; line-height: normal; font-size: 15px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">t=8st=8s
So we get an acceleration of:
a=−15m/s8s=−1.875m/s2" role="presentation" style="display: inline-table; font-style: normal; font-weight: normal; line-height: normal; font-size: 15px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">a=−15m/s8s=−1.875m/s2a=−15m/s8s=−1.875m/s2
When we have the acceleration, we can calculate the distance with the kinematic equation (for constant acceleration):
xf=x0+v0⋅t+12a⋅t2" role="presentation" style="display: inline-table; font-style: normal; font-weight: normal; line-height: normal; font-size: 15px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">xf=x0+v0⋅t+12a⋅t2xf=x0+v0⋅t+12a⋅t2
Where we calculate the total distance. We know every value on the right hand side, so we get:
xf=0+15m/s⋅8s+12(−1.875m/s2)⋅(8s)2=60m" role="presentation" style="display: inline-table; font-style: normal; font-weight: normal; line-height: normal; font-size: 15px; text-indent: 0px; text-align: center; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; width: 10000em !important; position: relative;">xf=0+15m/s⋅8s+12(−1.875m/s2)⋅(8s)2=60mxf=0+15m/s⋅8s+12(−1.875m/s2)⋅(8s)2=60m
So it takes the bus 60 meters to break at a constant acceleration as calculated above
54 km/h in seconds =
[tex]54 \times \frac{5}{18} [/tex]
= 15metere per second
acceleration formula = v-u upon t
u=0 v=15 t = 6
a=
[tex] \frac{v - u}{t} [/tex]
15 - 0 upon 6
[tex] \frac{15}{6} [/tex]
[tex] \frac{5}{2} [/tex]