Question #292238

A racing car is traveling along a horizontal track. The car, starting from rest accelerates uniformly at 4.50ms-2 for 15s when it reaches maximum speed. It then travels at this speed for 20s before deceleration uniformly at 5.60ms-2 until it come to rest.

I. Determine the maximum speed reached by the car

Ii. Determine the distance travelled by the car before reaching maximum speed

iii. Determine the time taken for the car to decelerate form it’s maximum speed and comes to rest

iv. Determine the total distance travelled by the car


Expert's answer

Given:

v0=0v_0=0

a1=4.50 m/s2a_1=4.50\rm \: m/s^2

t1=15 st_1=15\:\rm s

t2=20 st_2=20\:\rm s

a2=5.60 m/s2a_2=5.60\:\rm m/s^2

v2=0v_2=0


i. The maximum speed reached by the car

v1=v0+a1t1=0+4.50∗15=67.5 m/sv_1=v_0+a_1t_1=0+4.50*15=67.5\rm \: m/s

ii. The distance travelled by the car before reaching maximum speed

d1=v0t1+a1t12/2=0+4.50∗152/2=506.25 md_1=v_0t_1+a_1t_1^2/2=0+4.50*15^2/2=506.25\: \rm m

iii. The time taken for the car to decelerate form it’s maximum speed and comes to rest

t3=v1/a2=67.5/5.60=12 st_3=v_1/a_2=67.5/5.60=12\:\rm s

iv. The total distance travelled by the car

d=d1+v1t2+(v1+v2)/2t3=506.25+67.5∗20+67.5/2∗12=2263 md=d_1+v_1t_2+(v_1+v_2)/2t_3\\ =\rm 506.25+67.5*20+67.5/2*12=2263\: m


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