Brakes are applied to a moving truck to produce retardation of 5m/s2. If the time taken between application of brakes and the truck to stop is 2 seconds, calculate the distance travelled by the truck during this time
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In this question, you use the SUVAT equations:
v=u +at
s=ut+1/2at2
v2=u2+2as
s={(v+u)t}/2 where s-displacement
v-initial velocity
u-final velocity
a-acceleration/
decceleration
t-time
in this question we have:
t= 2s a=5m/s2 and, final velocity= om/s (after the car stops)
Therefore will use: s=ut + 1/2at2
s= 0 x 2 + 1/2 x 5 x 2(squared)
=0 + 1/2 x 5 x 4
=1/2 x 20
=10
therefore the distance covered = 10meters.
v=u +at
s=ut+1/2at2
v2=u2+2as
s={(v+u)t}/2 where s-displacement
v-initial velocity
u-final velocity
a-acceleration/
decceleration
t-time
in this question we have:
t= 2s a=5m/s2 and, final velocity= om/s (after the car stops)
Therefore will use: s=ut + 1/2at2
s= 0 x 2 + 1/2 x 5 x 2(squared)
=0 + 1/2 x 5 x 4
=1/2 x 20
=10
therefore the distance covered = 10meters.
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